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Definitions of important concepts and proof strategies for important results. These are things that you should make an effort to learn. Presented in vocabulary list format. $\DeclareMathOperator{\n}{null} \DeclareMathOperator{\r}{range}$

*1*Chapter 1¶

- Vector space
- A set, with addition (comm, assoc, identity, inverse, dist) and scalar mult (assoc, identity, dist)
- Subspace
- Subset. Contains 0, closed under addition, closed under scalar mult.
- Subspaces of $\mathbb R^2$
- Zero, lines passing through origin, all of $\mathbb R^2$
- Sum of subspaces
- Subspace, consisting of elements formed as a sum of vectors from each subspace. This is the smallest subspace containing all of the components.
- Direct sum
- Unique representation. Must be sum, and only 1 way to write 0 (i.e., intersection contains only zero).
- Intersection of subspaces
- always a subspace
- Union of subspaces
- only a subspace if one is contained within the other
- Addition on vector spaces
- comm, assoc, identity is $\{0\}$

*2*Chapter 2¶

- Span
- subspace. set of all linear combos
- Finite-dimensional vector space
- spanned by a finite list of vectors
- Linear independence
- A linear combo is 0 only when all the coefficients are 0
- one vector can be written as a linear combo of the others
- we can remove one without affecting the span
- Length of spanning sets and linearly independent lists
- Spanning sets are never shorter. Proof: replace vectors from a spanning list with vectors from a lin ind list, should still span, etc.
- Basis
- linearly independent spanning list
- any vector can be written as a unique combination of basis vectors
- any lin ind list can be reduced to a basis
- any spanning set can be extended to a basis
- Existence of a direct sum
- Given a subspace $U$ of $V$, there is a $W \subseteq V$ such that $U \oplus W = V$. Use the basis vectors, show $V$ is sum, intersection is 0 (from lin ind).
- Dimension of a vector space
- Number of vectors in a basis (any)
- Lunch in chinatown
- $\dim(U_1 + U_2) = \dim(U_1) + \dim(U_2) - \dim(U_1 \cap U_2)$; proof via bases (only holds for 2 subspaces)

*3*Chapter 3¶

- Linear map
- Function $T: V \to W$, satisfies additivity and homogeneity
- Assoc, dist, identity I; not comm
- Product of linear maps
- Composition, in the same order ($(ST)(v) = S(Tv)$)
- Nullspace
- Things in the domain that go to 0
- Always a subspace
- Injective $\Leftrightarrow$ nullspace only contains 0
- Range
- Whatever things in the domain map to
- Always a subspace
- Surjective $\Leftrightarrow$ range is $W$
- Rank-nullity theorem
- $\dim V = \dim \n T + \dim \r T$
- Proof: Create bases, apply T, nullspace parts disappear because they go to 0; this proves the spanning. To prove lin ind, use lin ind of the bases of the ambient space
- Corollary: no injective maps to a smaller VS, no surjective maps to a larger VS
- Matrix of a linear map
- Each column is $T$ applied to a basis vector (each row: coefficient for that basis vector)
- Invertibility of operators
- $T$ is invertible if $\exists \, S$ (unique) with $ST = TS = I$
- Invertibility $\Leftrightarrow$ bijectivity. Proof: show injectivity, surjectivity for one direction, and prove linearity of the inverse for the other
- Isomorphic vector spaces
- There is an invertible linear map (and thus bijection) between them
- Same dimension
- Dimension of the vector space of linear operators
- $\mathcal L(V, W) = \dim V \cdot \dim W$
- Operator
- Linear map from $V$ to $V$
- Proving direct sum
- To show that $U \oplus W = V$, show that their intersection is 0, and that you can make an arbitrary vector from $V$ by taking one from each
- Product of injective maps
- Also injective. Proof: apply argument inductively, starting from the end
- Injective maps and linear independence
- Applying an injective map on each element of a lin ind list gives a lin ind list. Proof: if a linear combo is 0, then apply $T$ gives 0, but nullspace is 0
- Invertibility of a product
- $ST$ is invertible $\Leftrightarrow$ $S, T$ both invertible. Proof: $T$ is inj, $S$ is surj; multiply to get $I$ in the other direction

*4*Chapter 4¶

- Roots of polynomials
- $p(\lambda) = 0$
- $p(z) = (z-\lambda)q(z)$. Proof: $p(z) - p(\lambda) = p(z) -0 $ and when you write it all out you can factor out $(z-\lambda)$ from each.
- At most $\deg p$ distinct roots
- Division algorithm for polynomials
- $q = sp + r$ where $\deg r < \deg p$
- Proof of uniqueness: assume two different representation, $(s-s')p = (r-r')$, look at degrees.
- Fundamental theorem of algebra
- Any polynomial has a unique factorisation into linear polynomials, with complex roots
- Unique factorisation of polynomials over the reals
- Linear and quadratic terms, all irreducible

*5*Chapter 5¶

- Invariant subspace
- If $u \in U$, then $Tu \in U$
- Means range is subspace of the domain
- Both range and nullspace are invariant for ops
- One-dimensional invariant subspaces
- Trivial (zero, whole space)
- Two-dimensional invariant subspaces
- Only the trivial ones for $\mathbb R$; $\mathbb C$ has some others
- Eigenvalue
- $\lambda$ such that $Tv = \lambda v$ for some $v \neq 0$
- $T$ has a one-dimensional invariant subspace
- $T-\lambda I$ is not injective/invertible/surjective (so it's 0 at some point)
- Eigenvectors
- In the nullspace of $T-\lambda I$ for an eigenvalue $\lambda$
- If every vector is an evector, then $T$ is $aI$ ($a$ is the only evector)
- Linear independence of eigenvector from distinct eigenvalues
- Proof: let $v$ be in the span of the previous, write it out as a linear combo and apply $T$ to get $\lambda_i$ as a coeff for each basis vector, multiply both sides by $\lambda$ and subtract from the above, then since $\lambda \neq \lambda_i$ and the LHS is 0, the coefficients must be 0.
- Number of eigenvalues
- Max: $\dim V$ since the eigenvectors are linearly independent
- Rotation
- No real eigenvalues - only complex (usually $\pm i$)
- Eigenvalues over a complex vector space
- Every operator has at least one. Proof: $(v, Tv, \ldots, T^nv)$ is not lin ind, so write 0 as a combo, turn this into a poly, factor over complex
- Upper triangular matrix
- Everything below the diagonal is 0
- Every op has one wrt some basis
- Means that $Tv_k$ for any basis vector is in the span of the previous ones
- Means that the span of the first $k$ basis vectors is invariant under $T$
- Zeroes and upper triangular matrices
- $T$ is invertible $\Leftrightarrow$ no zeroes on the diagonal of a UT matrix. Proof: $Tv = 0$ for some $v$, so not injective, so not invertible; other direction, write $Tv = 0$ as a combo of basis vectors, then $Tv_k$ is in span of previous, so the coeff of $v_k$ is 0
- Diagonal elements of upper triangular matrices
- These are eigenvalues. Proof: $T - \lambda I$ is not invertible iff there is a zero on the diagonal, which happens if $\lambda = \lambda_i$ for some $i$
- Diagonal matrices (TFAE)
- An operator has a diagonal matrix
- Evectors form a basis
- Sum of one-dimensional invariant subspaces is the whole space
- The sum of nullspaces of $T - \lambda_k I$ is the whole space
- Invariant subspaces on real vector spaces
- There is always one of dim 1 or 2 (proof by unique factorisation)
- Eigenvalues on odd-dimensional spaces
- Every op has one
- Invariance under every operator
- Then $U$ must be trivial (0 or V). Proof: Suppose there is nonzero $v \in U$, and $w \notin U$. Extend $v$ to a basis, map $av$ to $aw$, but that's not in $U$ so contradiction
- Eigenspace invariance if $ST = TS$
- Any eigenspace (or nullspace of $T-\lambda I$) of $T$ is invariant under $S$. Proof: If $v$ is in the eigenspace, then $(T-\lambda I)v = 0$. Apply to $Sv$, by distributivity we get 0, so $Sv$ is in eigenspace too, so the eigenspace is invariant
- Number of distinct eigenvalues
- Dimension of range + 1 if there is a 0 eigenvalue, just dim range otherwise. Proof: at most one from the nullspace (0)
- Eigenvalues of the inverse
- $1/\lambda$. Proof is easy
- Eigenvalues of $ST$ and $TS$
- The same
- Nullspace and range of $P^2=P$
- The direct sum is $V$. Proof: $u-Pv$ is in the nullspace, $Pv$ is in the range

*6*Chapter 6¶

- Inner product
- function that takes in $u, v \in V$, outputs something from the field
- Positive definiteness ($(\langle v, v \rangle \geq 0$), linearity in the first arg, conjugate symmetry, conjugate homogeneity in the second arg
- Standard inner products
- Euclidean (dot product with conjugate of second vector) on $\mathbb F^n$
- Integration for $p$
- Inner products and linear maps
- $f: v \to \langle v, w\rangle$ for fixed $w$ is a linear map (by the way IPs are defined) thus $\langle w, 0 \rangle = \langle 0, w \rangle = 0$
- Norm
- $\|v \| = \sqrt{\langle v, v \rangle}$
- Inherits positive definiteness (only 0 if $v$ is)
- $\|av\|^2 = |a|^2\|v\|^2$
- Orthogonal
- $\langle u, v \rangle = 0$
- 0 is orthogonal to everything, and is the only vector orthogonal to itself
- Pythagorean theorem
- If $u$ and $v$ are orthogonal, then $\|u+v\|^2 = \|u\|^2 + \|v\|^2$
- Proof: $\|u+v\|^2 = \langle u+v, u+v \rangle = \|u\|^2 + \|v\|^2 + \langle u, v \rangle + \langle v, u \rangle$ but the last two terms are 0 since they're orthogonal
- Orthogonal decomposition
- $v = au + w$ where $w$ is ortho to $u$. Set $w = v - au$, choose $a = \langle v, u \rangle / \|u\|^2$. Derivation: from $\langle v-au, u \rangle = 0$, rewrite in terms of norms
- Cauchy-Schwarz inequality
- $|\langle u, v \rangle | \leq \|u\|\|v\|$, equality if scalar multiple
- Proof: Assume $\|u\|^2 \neq 0$, divide by it, write ortho decomp of $v = u+w$, use Pythagorean to get $\|v\|^2$, multiply both sides by $\|u\|^2$, make an inequality; equality only if $w=0$
- Triangle inequality
- $\|u+v\| \leq \|u\| + \|v\|$
- Proof: Write it out, use Cauchy-Schwarz, equality if non-negative scalar mult
- Paralellogram inequality
- $\|u+v\|^2 + \|u-v\|^2 = 2(\|u\|^2+\|v\|^2)$
- Proof: from inner products
- Orthonormal list
- any two vectors are ortho, each has a norm of 1
- $\| a_1e_1 + \ldots + a_ne_n\|^2 = |a_1|^2 + \ldots + |a_n|^2$ by Pythagorean theorem
- Always linearly independent (use the previous thing, try to make 0)
- Linear combinations of orthonormal bases
- $v = \langle v, e_1\rangle e_1 + \ldots + \langle v, e_n \rangle$
- Proof: $v = a_1e_1 + \ldots$, take inner product of $v$ with each $e_j$, get $a_j$
- Also, $\|v\|^2 = |\langle v, e_1 \rangle |^2 + \ldots$
- Gram-Schmidt
- For creating orthonormal bases (for the same VS) out of lin ind lists
- $e_1 = v_1 / \|v_1\|$; $e_j = v_j-\langle v_j, e_1\rangle e_1 - \cdots$ then divide by the norm (not zero since the $v$s are lin ind)
- Proof: Norms are clearly 1. Orthogonality take the inner product $\langle e_j, e_k \rangle$, most terms disappear because of pairwise ortho, so we just have $\langle v_j, e_k \rangle - \langle v_j, e_k \rangle$. Also, $v_j$ is in the span (just rearrange the formulas) so they span the same space.
- Corollary: any FDIPS has an ortho basis
- If the list is linearly
*dependent*, we get a division by 0 - Upper triangular matrices and orthonormal bases
- If an op has a UT matrix wrt some basis, it has one wrt to an ortho one (so in a complex VS, this is every op)
- Proof: span of $(v_1, \ldots, v_j)$ is invariant under $T$ for each $j$, apply Gram-Schmidt to get an ortho basis
- Orthogonal complement
- $U^{\perp}$ = set of all vectors orthogonal to every vector in $U$
- Is a subspace
- $(U^{\perp})^{\perp} = U$ for obvious reasons
- Direct sum of orthogonal complement
- $U \oplus U^{\perp} = V$
- Proof: use an ortho basis for $U$. $v = (\langle v, e_1\rangle + \ldots) + (v -\langle v, e_1\rangle - \ldots) = u + w$. Clearly $u \in U$, and $w\in U^{\perp}$ because $\langle w, e_j\rangle = 0$ (so $w$ is ortho to every basis vector of $U$). Intersection is obviously only 0, by positive definiteness
- Orthogonal projections
- $P_Uv$ maps $v$ to the part of its ortho decomp that is in $U$
- Range is $U$, nullspace is $U^{\perp}$
- $v-P_Uv$ is in the nullspace
- $P_U^2 = P_U$
- $\|P_Uv \| \leq \|v \|$
- Minimisation problems
- Find $u \in U$ to minimise $\|v-u\|$ for fixed $v$. Answer: $u = P_Uv$!
- Proof: $\|v-P_Uv \|^2 \leq \|v-P_Uv\|^2 + \|P_Uv-u\|^2 = \|v-u\|^2$ by Pythagorean theorem (applicable since vectors are ortho; middle terms cancel out), equality only when $u = P_Uv$
- First, find an ortho basis for the subspace we're interested in (e.g., polynomials), take inner product of $v$ with each basis vector and use as coefficient
- Linear functional
- $\varphi: V \to \mathbb F$ (so sending $v$ to $\langle v, u\rangle$ for fixed $u$)
- Existence of $u$: write $v$ in terms of ortho basis, use homogeneity, find $u$
- Uniqueness of $u$: assume $\langle v u_1, \rangle = \langle v, u_2 \rangle$ so $0 = \langle v, u_1-u_2\rangle$ for any $v$ including $u_1 - u_2$ thus $u_1-u_2 = 0$ QED
- Adjoint
- If $T \in \mathcal L(V, W)$, $T^* \in \mathcal L(W, V)$ such that $\langle Tv, w \rangle = \langle v, T^*w \rangle$
- Always a linear map
- Eigenvalues are conjugates of $T$'s eigenvalues (proof by contradiction: $T-\lambda I \neq 0$, so there's an inverse, apply adjoint op to both sides, also invertible, thus not an eigenvalue)
- If injective, original is surjective, etc (all 4 possibilities)
- The adjoint operator
- The operator analogue of [conjugate] matrix transposition
- Additivity ($(S+T)^* = S^* + T^*$), conjugate homogeneity ($(aT)^* = \overline{a}T^*$)
- $(T^*)^* = T$, $I^* = I$
- $(ST)^* = T^*S^*$
- Nullspace of $T^*$ is complement of range of $T$, and range is complement of nullspace of $T$, and the other way around

*7*Chapter 7¶

- Self-adjoint operator (Hermitian)
- $T^* = T$, matrix is equal to conjugate transpose (ONLY WRT AN ORTHO BASIS)
- Preserved under addition, real scalar mult
- All eigenvalues are real (proof: definitions and conj symmetry; $\lambda \|v\|^2 = \overline{\lambda}\|v\|^2$)
- $\langle Tv, v \rangle \in \mathbb R$ (proof: subtract conjugate, use conj symmetry, zero operator thing below)
- Product of two self-adjoint ops only self-adjoint if the multiplication is commutative
- Set of all self-adjoint ops is a subspace only in a real IPS, not a complex IPS
- Orthogonal projections are self-adjoint
- Zero operators on complex inner product spaces
- If $\langle Tv, v \rangle = 0$ for all $v$ then $T = 0$
- Normal operator
- $T^*T = TT^*$
- May not be self-adjoint
- $\lVert Tv\rVert = \lVert T^*v\rVert$. Proof: $\langle (TT^*-T^*T)v, v \rangle = 0$ so $\langle TT^*v, v \rangle = \langle T^*Tv , v \rangle$ then use adjoint def to get $\langle T^*v, T^*v \rangle = \langle Tv, Tv \rangle$
- Eigenvectors for $\lambda$ are eigenvectors for $\overline{\lambda}$. Proof: $(T-\lambda I)$ is normal, times $v$ is 0, use norm relation above to get $\|(T-\lambda I)^*v \| =0$, distribute the adjoint
- Set of all normal ops on a VS with $\dim >1$ is not a subspace (additivity not satisfied)
- $T^k, T$ have the same range and nullspace
- Every normal op on a complex IPS has roots
- Orthogonality of eigenvectors of normal operators
- Evectors for distinct evalues are ortho. Proof: $(\lambda_1 - \lambda_2)\langle v_1, v_2\rangle = \langle Tv_1, v_1 \rangle - \langle v_2, T^*v_2 \rangle = 0$ but evalues are distinct so inner product must be 0.
- Spectral theorem
- For which ops can evectors form an ortho basis (or have diagonal matrices wrt to an ortho basis)
- Complex spectral theorem
- Normal $\Leftrightarrow$ eigenvectors form an ortho basis
- Proof: Diagonal matrix wrt a basis, conjugate transpose also diagonal, so they commute, so $T$ is normal. Other direction: $T$ has a UT matrix, this is diagonal if you look at the sum of squares in the $j$th row versus $j$th column and use induction (and the fact that $\|Te_j \| = \|Te^*e_j\|$)
- Real spectral theorem
- Self-adjoint $\Leftrightarrow$ eigenvectors form an orthonormal basis
- Lemma: self-adjoint op has an eigenvalue in a real IPS. Proof: consider $(v, Tv, \ldots, T^nv)$, not lin ind, write 0, factor over the reals, none of the quadratics is 0, so one of the linears is 0
- Proof: Induction. $T$ has at least one evalue and evector $u$, which is the basis for a 1-d subspace $U$. $U^{\perp}$ is invariant under $T$ (reduces to $\lambda \langle u, v \rangle = 0$). Create a new op $S|_{U^{\perp}}$, which is self-adjoint, apply the IH, join it with $u$
- Normal operators on two-dimensional spaces (TFAE)
- $T$ is normal but not self-adjoint
- Matrix wrt any ortho basis looks like $\displaystyle \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ but not diagonal. Proof: $\|Te_1 \|^2 = \|T^*e_1\|^2$ to find $c$, use the fact that $T$ is normal to do matrix mult and find $d$
- For
*some*ortho basis, $b < 0$ (above). Proof: use $(-e_1, -e_2)$ as the basis - Block matrix
- When an entry of a matrix is itself a matrix
- Invariant subspaces and normal operators
- $U^{\perp}$ is also invariant under $T$. Proof: consider basis vectors, and the matrix, which has all zeros under the first few columns since $U$ is invariant, but since $\|Te_j\|^2 = \|T^*e_j\|$ then yeah
- $U$ is invariant under $T^*$ (take transpose of above matrix)
- Block diagonal matrix
- Square matrix, diagonal consists of block matrices (of any form), all others are 0 (includes all square matrices really)
- Product of two block diagonal matrices: multiply the matrices together along the diagonal, stick the result where you expect
- Block diagonal matrices of normal operators
- Normal $\Leftrightarrow$ has a block diagonal matrix wrt some ortho basis, each block is 1x1 or 2x2 of the form $\displaystyle \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ with $b > 0$. Proof by strong induction, eigenvalues, invariant subspaces to form ortho bases, if dim is 2 then $T|_U$ is normal but not self-adjoint, apply IH to $U^{\perp}$ and join it with the ortho basis of $U$
- Generalised eigenvector
- $(T-\lambda I)^jv =0$ for some $j > 0$
- Used to write $V$ as the decomp of nullspaces (generalised eigenspaces basically), where $j = \dim V$
- Nullspaces of powers $T^0, T^1, T^2, \ldots$
- Size of nullspace always monotonically increasing
- If two consecutive nullspaces are equal, all subsequent ones are equal too (proof: consider nullspace of $T^{m+k+1}$, so $T^kv$ is in nullspace of $T^{m+1}$, so it's in the nullspace of $T^{m}$ too
- $T^{\dim V}$ and $T^{\dim V+1}$ have the same nullspace etc which is how we get $\dim V$ for $j$ above
- Nilpotent operators
- $N^k = 0$ for some $k$
- $k \leq \dim V$ since every $v$ is a generalised eigenvector for $\lambda = 0$ so it inherits it from the previous statements
- $(v, Tv, Tv^2, \ldots, T^mv)$ is lin ind if $T^mv \neq 0$ (try to make 0, apply $T$ to both sides, all coeffs are 0)
- If $ST$ is nilpotent, so is $TS$ (proof: $(TS)^{n+1} = 0$; regroup etc)
- Only $N=0$ is self-adjoint and nilpotent
- If 0 is the only evalue on a CVS, $N$ is nilpotent, because every vector is a generalised evector
- Ranges of powers
- Size monotically decreasing
- $T^{\dim V}$ and $T^{\dim V+1}$ have the same range
- Multiplicity
- Means geometric multiplicity (dimension of eigenspace)
- Controls the number of times an eigenvalue shows up on the diagonal
- Sum of multiplicities if $\dim V$ on a complex VS
- Characteristic polynomial
- Factors of eigenvalues, algebraic multiplicity. Degree is $\dim V$. Roots are eigenvalues.
- Cayley-Hamilton
- Char poly is $q(z)$, then $q(T) = 0$
- Proof: Show $q(T)v_i = 0$ for every basis vector $v_i$. Strong induction on dim. For $n=1$, obvious, only factor vanishes. For $n$, from the UT matrix we can tell that $v_i$ is in the span of the previous $v$s, so we write it as a linear combo, and apply the IH (so all the other factors go to 0)
- Nullspaces of polynomials of $T$
- Invariant under $T$ somehow
- Decomposition into nilpotent operators
- The generalised eigenspaces
- Bases from generalised eigenvectors
- Always enough to form a basis for a complex VS
- Matrices of nilpotent operators
- There's always a UT matrix with 0's along the diagonal wrt some basis (choose bases from the nullspace of $N$, then $N^2$, put them all together in one giant basis)
- Upper-triangular block matrices
- If $T$ has $m$ distinct eigenvalues, $T$ has a block diagonal matrix where each block is UT with an eigenvalue repeated along the diagonal, the number of times acc. to its geometric mult
- Square roots of operators
- If $N$ is nilpotent $I+N$ has a square root (and any other root); proof by Taylor expansion, which is a finite sum because $N^m = 0$ for all $m$ after something
- Any invertible op has a sqrt on a complex VS - eah eigenspace has a nilpotent op ($\lambda I + N$) and since $T$ is invertible, none of the eigenvalues are 0, so we can divide by $\lambda$, showing that $S = \lambda I + N$ which is just $T$ limited to one subspace, then extend this to the ambient space
- Minimal polynomial
- The unique monic poly $p$ of smallest degree so that $p(T) = 0$
- Existence proof: $(I, T, T^2, \ldots T^{n^2})$ is lin dep (since $\dim \mathcal L(V) = n^2$) so make 0 using some choice of coeffs, not all zero
- Unique if we limit to the smallest degree (so instead of $n^2$ use the smallest $m$ that makes it dep)
- $q(T) = 0$ $\Leftrightarrow$ min poly divides $q$ (for any $q$, including the char poly); proof by division algo
- Roots are eigenvalues, proof by non-zeroness of eigenvectors
- All factors of the char poly are in the min poly, multiplicity is reduced to 1 only if all the geometric and algebraic multiplicities are the same (so eigenvectors form a basis)
- Nilpotent operators and bases
- We can make a basis out of $(v_1, Nv_1, \ldots, N^{a_1}v_1, v_2, \ldots)$ (the $a$'s are the highest power that don't make it 0)
- Proof: strong induction on dim, apply IH on range since nullspace is not just zero (cuz nilpotent), make a basis out of range vectors, then consider complement of nullspace within range
- Jordan form
- block-diagonal matrix, each block has size determined by geometric multiplicity and has the evalue along the diagonal and 1 in the line above it
- Existence proof for $T$ in a CVS: works for nil ops $(N^{a_1}v_1, \ldots, Nv_1, v_1)$ giving us zero along the diagonal, and eigenspaces are given by nil ops etc
- The matrix of $(v_n, \ldots, v_1)$ has each block transposed, so the diagonal is flipped along the / axis (still a diagonal, in reverse order) - so the 1s are under the diagonal

*8*Chapter 9¶

Omitted (not covered)

*9*Chapter 10¶

- Change-of-basis matrix
- To convert from one basis to another, figure it out from first principles I guess
- Equivalent to the matrix for the op that maps each basis vector to the corresponding one
- Inverse matrix gives the opposite direction
- Trace
- Independent of the basis
- On a CVS, equal to sum of eigenvalues (repeated acc. to geometric multiplicity)
- On a RVS, sum of eigenvalues minus sum of first coordinates of eigenpairs (again, geo mult)
- sum along the diagonal for a UT matrix (same as sum of evalues), or even a non-UT matrix
- $BA = AB$ (square, same size)
- No operators such that $ST - TS = I$, by looking at trace
- Is a non-negative int if $P^2=P$
- trace of $(T^*T) = \|Te_1\|^2 + \ldots$ (cuz trace is given by $\langle T^*T e_1, e_1 \rangle$, since that's how the matrix works etc)
- Determinant
- $(-1)^{\dim V}$ times the constant term in the char poly
- CVS: product of eigenvalues (incl repeats, acc. to geo mult)
- RVS: product of evalues and second coordinates of eigenpairs (geo mult)
- Invertible op $\Leftrightarrow$ non-zero determinant. Proof: det is zero iff an eigenvalue is 0, in which case, not invertible
- Char poly is $\det(zI-T)$
- For a diagonal matrix, just take the product of the diagonal elements
- For block UT matrices, the det is the prod of the det of the block matrices along the diagonal
- Changing two columns flips the sign (permutation theory)
- If a column is a scalar mult of another, the det is 0
- $\det(AB) = \det(BA) = \det(A)\det(B)$
- Dets of ops are independent of the basis used for the matrix