Math 110 Linear Algebra Midterm 2

Take arbitrary $p, q$ and a scalar $\lambda$. Show $T(p + q) = T(p) + T(q)$ AND $T(\lambda p) = \lambda T(p)$. Start each line from the left side.

Not linearity. That only says the outputs land in $\mathbb{R}^2$. Linearity compares $E(p_1 + p_2)$ with $E(p_1) + E(p_2)$, and $E(\lambda p)$ with $\lambda E(p)$.

Write $p = a_0 + a_1x + a_2x^2 + a_3x^3$. Every condition on $p$ (like $p(1) = 0$) becomes a linear equation in $a_0, \dots, a_3$.

General $p$, impose $Tp = 0$, solve for the $a_i$, write $p$ as (free variable) times (fixed polynomial) + ... Those fixed polynomials are the basis. Count-check with rank-nullity first.

For an arbitrary target $(a_1, \dots, a_n)$, name an input that maps to it. Or show $\dim \operatorname{range} T = n$.

That's a description, not a basis. A basis is a LIST $u_1, \dots, u_k$ of vectors in $U$ that is LI and spans $U$.

$\dim V = \dim \operatorname{null} T + \dim \operatorname{range} T$ ($V$ finite-dim).

$T$ is injective iff $\operatorname{null} T = \{0\}$.

$A = v + U = \{v + u : u \in U\}$ for some $v \in V$ and subspace $U$. A subspace shifted off the origin.

The set of all affine subsets $v + U$ parallel to $U$, with $(v + U) + (w + U) = (v + w) + U$ and $\lambda(v + U) = \lambda v + U$.

$0 + U = U$. So $v + U$ is zero in $V/U$ exactly when $v \in U$.

(i) $v - w \in U$. (ii) $v + U = w + U$. (iii) $(v + U) \cap (w + U) \neq \emptyset$. Two cosets are equal or disjoint.

Check $v - w \in U$ (Prop 39).

A coset has many names: $v_1 + U = v_2 + U$ with $v_1 \neq v_2$. The answer can't depend on which name you picked.

Start from the hypothesis $v + U = w + U$. Prop 39 gives $v - w \in U$. End at: the formula gives the same output for $v$ and $w$.

$\pi : V \to V/U$, $\pi(v) = v + U$. Linear, surjective, and $\operatorname{null} \pi = U$.

$\dim V/U = \dim V - \dim U$ (finite-dim). Proof: rank-nullity on $\pi$.

$\overline{T} : V/\operatorname{null} T \to \operatorname{range} T$, $\overline{T}(v + \operatorname{null} T) = Tv$, is an isomorphism, and $T = \iota \circ \overline{T} \circ \pi$.

1 well-defined (via Prop 39). 2 linear. 3 injective (output 0 means $v \in \operatorname{null} T$, so $v + \operatorname{null} T$ is the zero coset). 4 surjective (every $Tv$ is hit by $v + \operatorname{null} T$).

Take a basis $u_1, \dots, u_m$ of $U$. Extend it to a basis of $V$ with $w_1, \dots, w_k$. Then $w_1 + U, \dots, w_k + U$ is a basis of $V/U$.

If $T(U_1) \subseteq U_2$, then $\overline{T}(v + U_1) = Tv + U_2$ is a well-defined linear map $V/U_1 \to W/U_2$.

$T(U_1) \subseteq U_2$. If $v - w \in U_1$, then $Tv - Tw = T(v - w) \in U_2$, so $Tv + U_2 = Tw + U_2$.

$U/(U \cap W) \cong (U + W)/W$, via $u + (U \cap W) \mapsto u + W$.

If $W \subseteq U \subseteq V$, then $(V/W)/(U/W) \cong V/U$.

$D : \mathcal{P}(\mathbb{R}) \to \mathcal{P}(\mathbb{R})$ is surjective with null space the constants $\mathbb{R}$. So $\mathcal{P}(\mathbb{R})/\mathbb{R} \cong \mathcal{P}(\mathbb{R})$ via $p + \mathbb{R} \mapsto p'$. The inverse is the indefinite integral.

Yes. $(1,2) - (1,4) = (0,-2) \in U$. The quotient can't see the $y$-coordinate.

False. $v + U = 0$ whenever $v \in U$.

True (Prop 40).

False unless $U = \{0\}$. Its null space is $U$.

$V' = \mathcal{L}(V, \mathbb{F})$: all linear maps $\varphi : V \to \mathbb{F}$ (linear functionals).

A linear measurement of a vector. $\varphi(v)$ is one number about $v$.

$\dim V' = \dim V$ when $V$ is finite-dim.

For a basis $v_1, \dots, v_n$ of $V$: $\varphi_j(v_k) = \delta_{jk}$, which is 1 if $j = k$ and 0 otherwise.

Reads off the $j$-th coordinate: $\varphi_j(v) = a_j$.

The dual basis $\varphi_1, \dots, \varphi_n$ is a basis of $V'$.

$\dim V' = \dim V = n$ and the list has length $n$. Right length + LI means basis.

Suppose $\lambda_1\varphi_1 + \dots + \lambda_n\varphi_n = 0$ (the zero functional). Plug in $v_k$: the left side gives $\lambda_k$, so every $\lambda_k = 0$.

The row vectors $(1, 0, \dots, 0), \dots, (0, \dots, 0, 1)$. Functionals on $\mathbb{F}^n$ are row vectors.

False. It depends on the whole basis. For $(1,0), (0,1)$: $\varphi_1(x,y) = x$. For $(1,0), (1,1)$: $\varphi_1(x,y) = x - y$.

For $T : V \to W$: $T' : W' \to V'$, $T'(\varphi) = \varphi \circ T$. Do $T$, then measure.

$(T'\varphi)(v) = \varphi(T(v))$. Plug in a general $v$ and simplify. The answer is a functional on $V$ (a formula or row vector).

$T^{-1}$ eats vectors in $W$. $T'$ eats functionals on $W$ and returns functionals on $V$. $T'$ exists for every linear $T$, invertible or not.

Type error. $T'$ only takes functionals. Write $(T'\varphi)(v)$.

False. $T' : W' \to V'$. The direction flips.

False. Every linear map has a dual map.

$T'(\varphi)(p) = \varphi(xp) = 2p(2)$, so $T'\varphi = 2\varphi$.

$(T'\ell)(x, y) = \ell(x + 3y) = 2x + 6y$. Row vector $(2, 6)$.

Show they give the same number at every $v \in V$ (or at every vector of a basis).

$T \mapsto T'$ is linear: $(\lambda T_1 + T_2)' = \lambda T_1' + T_2'$. And it reverses products: $(ST)' = T'S'$.

$(ST)'\varphi = \varphi \circ S \circ T = (S'\varphi) \circ T = T'(S'\varphi)$. $S'$ acts first.

In the dual bases, $\mathcal{M}(T')$ is the transpose of $\mathcal{M}(T)$.

$\mathcal{M}(T)$ is $2 \times 3$. $\mathcal{M}(T')$ is $3 \times 2$.

$A' : (\mathbb{R}^2)' \to (\mathbb{R}^3)'$, $(x\ \ y) \mapsto (x\ \ y)A$. Its matrix is $A^T$, rows $(1, 4), (2, 5), (3, 6)$.

$(AB)^T = B^TA^T$.

$U^0 = \{\varphi \in V' : \varphi(u) = 0 \text{ for all } u \in U\}$. The measurements that read 0 on all of $U$.

$\dim U + \dim U^0 = \dim V$ (finite-dim).

Look for row vectors $c$ with $c \cdot u_i = 0$ for every vector $u_i$ in a basis of $U$. Solve the system.

True.

$\operatorname{null} T' = (\operatorname{range} T)^0$ and $\operatorname{range} T' = (\operatorname{null} T)^0$.

$T$ injective iff $T'$ surjective. $T$ surjective iff $T'$ injective.

Prop 23 on $T'$: suppose $T'\varphi = 0$. Unfold Def 28 at every $v$. Use surjectivity to reach every $w \in W$.

False. You may quote results unless the problem asks you to prove them.

$j_V(v)(\varphi) = \varphi(v)$ defines an injective linear map $V \to V''$. It's an isomorphism when $V$ is finite-dim.

If $\varphi(v) = \varphi(w)$ for every $\varphi \in V'$, then $v = w$. A vector is determined by all its linear measurements.

$T'' \circ j_V = j_W \circ T$ for every linear $T : V \to W$.

True. Its range is a subspace of $\mathbb{F}$, so it's $\{0\}$ or $\mathbb{F}$.

$U$ is invariant under $T$ if $T(U) \subseteq U$.

Check $Tu_i \in U$ for each vector $u_i$ of a basis of $U$.

$\{0\}$, $V$, $\operatorname{null} T$, $\operatorname{range} T$.

$\lambda$ is an eigenvalue if $Tv = \lambda v$ for some $v \neq 0$. That $v$ is an eigenvector.

$E(\lambda, T) = \operatorname{null}(T - \lambda I)$.

Is $\operatorname{null}(T - \lambda I) \neq \{0\}$? Same as: $T - \lambda I$ is not injective.

Backwards. $v = 0$ always solves $Tv = \lambda v$. Only-zero means $\lambda$ is NOT an eigenvalue.

Every operator on a nonzero finite-dim COMPLEX vector space has an eigenvector.

Eigenvectors for distinct eigenvalues are linearly independent.

An operator on $V$ has at most $\dim V$ distinct eigenvalues.

None over $\mathbb{R}$. Over $\mathbb{C}$: $i$ and $-i$.

$t^2 - (a + d)t + (ad - bc)$. Its roots are the eigenvalues.

$T$ commutes with $T - \lambda I$.

True.

$a_{ij} = 0$ for $i > j$: every entry below the diagonal is 0.

$\mathcal{M}(T)$ is upper-triangular iff $Tv_j \in \operatorname{Span}(v_1, \dots, v_j)$ for each $j$ iff each $\operatorname{Span}(v_1, \dots, v_j)$ is invariant.

If $\mathcal{M}(T)$ is upper-triangular, the eigenvalues of $T$ are exactly its diagonal entries.

Over $\mathbb{C}$ ($V \neq 0$, finite-dim), some basis makes $\mathcal{M}(T)$ upper-triangular.

Induction on $\dim V$. Base: $\dim V = 1$. Step: eigenvalue $\lambda$ (Prop 53), $U = \operatorname{range}(T - \lambda I)$, invariant with $\dim U < \dim V$, IH on $T|_U$, extend to a basis of $V$, $Tv_j = (T - \lambda I)v_j + \lambda v_j$, finish with Prop 59.

$\lambda$ is an eigenvalue, so $\operatorname{null}(T - \lambda I) \neq \{0\}$. Then rank-nullity.

The parts ARE the proof skeleton. Write one line per part before deciding to skip. Part (a) is usually cheap points.

Base case (smallest $n$). Inductive hypothesis (assume for all smaller dims / for $n$). Step (show for the next one, and point at the line that uses the IH).

Some basis makes $\mathcal{M}(T)$ diagonal. Same as: $V$ has a basis of eigenvectors.

Diagonalizable. Eigenbasis exists. $V = E(\lambda_1, T) \oplus \dots \oplus E(\lambda_m, T)$. $\dim V = \sum \dim E(\lambda_i, T)$.

$\dim V$ distinct eigenvalues implies diagonalizable. Not the converse.

In an eigenbasis, $T^k(a_1v_1 + \dots + a_nv_n) = a_1\lambda_1^kv_1 + \dots + a_n\lambda_n^kv_n$.

$\dim E(\lambda, T)$.

False. $I$ is diagonal with one eigenvalue.

False. Only eigenvalue 1 and $\dim E(1) = 1 < 2$.

False. Rotation $R(x, y) = (-y, x)$.

If $p(x) = a_nx^n + \dots + a_1x + a_0$, then $p(T) = a_nT^n + \dots + a_1T + a_0I$.

The monic polynomial of smallest degree with $m_T(T) = 0$.

Find the first $k$ with $T^k \in \operatorname{Span}(I, T, \dots, T^{k-1})$. Solve $T^k = c_0I + \dots + c_{k-1}T^{k-1}$ by matching entries. Then $m_T = x^k - c_{k-1}x^{k-1} - \dots - c_0$.

Ask "is $T = cI$?", not "is $T = cT$?". The degree-2 test is $T^2 = aT + bI$.

$q(T) = 0$ iff $m_T$ divides $q$.

The eigenvalues of $T$ are exactly the zeros of $m_T$ in $\mathbb{F}$.

$\deg m_T \le \dim V$.

$T$ is diagonalizable iff $m_T = (x - \lambda_1) \cdots (x - \lambda_m)$ with distinct $\lambda_i \in \mathbb{F}$.

$T$ is invertible iff $m_T(0) \neq 0$ (the constant term isn't 0).

Start from $m_T(T) = 0$ with constant term $a_0 \neq 0$. Move $a_0I$ to one side, factor $T$ out of the rest, divide by $-a_0$.

No. Over $\mathbb{R}$ there may be none, and 0 can be one. Invertible iff 0 is NOT an eigenvalue.

$m_T$ is a polynomial, not a map. The inverse is $T^{-1}$, written as a polynomial in $T$.

$x - 1$ (Example 58).

$x^2$.

$x^2 + 1$, since $R^2 = -I$ and $R$ is not a multiple of $I$.

False. $I$ on $\mathbb{C}^2$: $m = x - 1$, characteristic $(x - 1)^2$.

True. $m_T$ divides $x^3 - 1$, which has 3 distinct roots in $\mathbb{C}$ (Props 65, 68).

True (Prop 66).

$G(\lambda, T) = \bigcup_k \operatorname{null}(T - \lambda I)^k$. Its nonzero vectors are generalized eigenvectors.

$\{0\} = \operatorname{null} T^0 \subseteq \operatorname{null} T^1 \subseteq \operatorname{null} T^2 \subseteq \dots$, and once $\operatorname{null} T^m = \operatorname{null} T^{m+1}$, it stays the same forever.

$G(\lambda, T) = \operatorname{null}(T - \lambda I)^{\dim V}$.

$G(\lambda, T) \neq \{0\}$ iff $\lambda$ is an eigenvalue.

Over $\mathbb{C}$: $V = G(\lambda_1, T) \oplus \dots \oplus G(\lambda_m, T)$.

$\dim G(\lambda, T)$.

$\dim E(\lambda, T) \le \dim G(\lambda, T)$. Diagonalizable iff they're equal for every eigenvalue.

Over $\mathbb{C}$, with distinct eigenvalues $\lambda_i$ and algebraic multiplicities $d_i$: $q(x) = (x - \lambda_1)^{d_1} \cdots (x - \lambda_m)^{d_m}$ satisfies $q(T) = 0$.

$N^k = 0$ for some $k$.

$N^{\dim V} = 0$, and 0 is its only eigenvalue.

True (Prop 73).

False. In $J$ with rows $(0, 1), (0, 0)$, $e_2$ is a generalized eigenvector for 0 but $Je_2 = e_1 \neq 0$.

Square block with $\lambda$ on the diagonal, 1s just above the diagonal, 0 elsewhere. The $\lambda$'s of different blocks don't have to be distinct.

Over $\mathbb{C}$, every operator has a basis in which its matrix is in Jordan normal form.

Number of Jordan blocks for $\lambda$ of size exactly $k$ = $2\dim\operatorname{null}(T - \lambda I)^k - \dim\operatorname{null}(T - \lambda I)^{k-1} - \dim\operatorname{null}(T - \lambda I)^{k+1}$.

$\dim\operatorname{null}(T - \lambda I)^k - \dim\operatorname{null}(T - \lambda I)^{k-1}$.

Number of blocks for $\lambda$ = geometric multiplicity. Total size of those blocks = algebraic multiplicity. Largest block size = exponent of $(x - \lambda)$ in $m_T$.

Characteristic $(x - 3)^3$. Minimal $(x - 3)^2$. Geometric 2, algebraic 3, so not diagonalizable.

True.

Only just above the diagonal, inside a block. Never between two blocks.