Linear Algebra

The first condition is linear: if $f(0)=2f(1)$ and $g(0)=2g(1)$, then $(f+\lambda g)(0)=2(f+\lambda g)(1)$; and 0 satisfies it, as do sums of continuous functions. The second fails at 0: $0\ne0+1$.

Write $f=\frac{f(x)+f(-x)}2+\frac{f(x)-f(-x)}2$: even plus odd. If $f$ is both even and odd, $f(x)=f(-x)=-f(x)$, so $f=0$; by Axler 1.46 the sum is direct.

Check the three conditions; each holds in both subspaces separately.

$0\in V_1$ and $0\in V_2$, so $0\in V_1\cap V_2$. If $u,w\in V_1\cap V_2$, then $u+w$ is in $V_1$ (closed under addition) and in $V_2$, so in the intersection. The same for $\lambda u$. The argument works for the intersection of any collection of subspaces, which is why "the smallest subspace containing a set" (its span) makes sense.

If neither contains the other, pick $u\in U\setminus W$ and $w\in W\setminus U$ and ask where $u+w$ lives.

If $U\subseteq W$, the union is $W$, a subspace; likewise if $W\subseteq U$.

Conversely, suppose $U\cup W$ is a subspace but neither contains the other. Take $u\in U\setminus W$ and $w\in W\setminus U$. Then $u+w\in U\cup W$. If $u+w\in U$, then $w=(u+w)-u\in U$, a contradiction. If $u+w\in W$, then $u=(u+w)-w\in W$, a contradiction. (The axes in the figure of this module are the picture.)

$U$ is 2-dimensional, so $W$ should be too, and should meet $U$ only in 0. Try fixing coordinates 2 and 4.

Take $W=\{(0,s,0,t):s,t\in\mathbf F\}$. Every vector splits: $(a,b,c,d)=(a,a,c,c)+(0,b-a,0,d-c)$. And if $(x,x,y,y)=(0,s,0,t)$, then $x=y=0$, so $U\cap W=\{0\}$. By Axler 1.46 the sum is direct.

There are many other choices: any 2-dimensional $W$ with $U\cap W=\{0\}$ works.

Add $e_1,e_2,e_3$ and discard dependent vectors. $e_1=(1,0,0)$ is not in the span (a combination $a(1,1,0)+b(0,1,1)$ has middle entry $a+b$ and first entry $a$, last $b$; $a=1,b=0$ gives middle 1). So $(1,1,0),(0,1,1),(1,0,0)$ is independent of length 3, hence a basis.

Extend a basis $u_1,\dots,u_m$ of $U$ to a basis $u_1,\dots,u_m,w_1,\dots,w_n$ of $V$, and let $W=\mathrm{span}(w_j)$. Every $v$ splits, and $U\cap W=0$ by independence. Not unique: in $\R^2$, any line other than $U$ itself is a complement of a line $U$. An orthogonal complement (module 5) picks a canonical one.

The plane is 2-dimensional. Any two independent vectors in it span it, and extra vectors in it don't enlarge the span.

Take $(1,-1,0)$, $(1,0,-1)$, $(0,1,-1)$, $(2,-1,-1)$. Each lies in the plane. The first two are independent (look at the last two coordinates), so they span a 2-dimensional subspace of the 2-dimensional plane, hence all of it. Adding the other two, which lie in the plane, doesn't change the span. (The plane is 2-dimensional because it is the null space of the nonzero map $(x,y,z)\mapsto x+y+z$ onto $\mathbf F$, by the fundamental theorem of linear maps.)

The first two vectors are independent, so the list is dependent iff the third is a combination of the first two.

$(2,3,1)$ and $(1,-1,2)$ are not multiples of each other, so they are independent, and by the linear dependence lemma the list is dependent iff $(7,3,c)=a(2,3,1)+b(1,-1,2)$ for some $a,b$. The first two coordinates give $2a+b=7$ and $3a-b=3$, so $a=2$, $b=3$. The third coordinate then forces $c=a+2b=8$.

Apply $\dim(U\cap W)=\dim U+\dim W-\dim(U+W)$ twice, using $\dim(U+W)\le10$.

$\dim(V_1\cap V_2)\ge7+7-10=4$. Then $\dim\big((V_1\cap V_2)\cap V_3\big)\ge4+7-10=1$. So the triple intersection contains a nonzero vector.

First: yes, uniquely, since $(1,1),(1,-1)$ is a basis; it is $T(x,y)=(x+y,x-y)$. Second: no, since $(2,4)=2(1,2)$ would force $T(2,4)=(2,0)$.

$\mathrm{range}(ST)\subseteq\mathrm{range}\,S$. Also $\mathrm{range}(ST)=S(\mathrm{range}\,T)$, the image of a space of dimension $\dim\mathrm{range}\,T$ under a linear map, so its dimension is at most that (fundamental theorem applied to $S$ restricted to $\mathrm{range}\,T$).

The fundamental theorem says the domain must be 5-dimensional.

$T:\mathbf F^5\to\mathbf F^2$, $T(x_1,\dots,x_5)=(x_1,x_2)$. Its null space is $\{(0,0,x_3,x_4,x_5)\}$, of dimension 3, and it is onto $\mathbf F^2$.

Find two such maps whose sum has a small null space.

Let $S(x)=(x_1,x_2,0,0)$ and $T(x)=(0,0,x_3,x_4)$. Both have null spaces of dimension 3. But $(S+T)(x)=(x_1,x_2,x_3,x_4)$ has null space $\{(0,0,0,0,x_5)\}$, of dimension 1. The set is not closed under addition.

Write an arbitrary vector of the range as $Tv$ and expand $v$.

Each $Tv_k\in\mathrm{range}\,T$, so their span is inside the range. Conversely, any $w\in\mathrm{range}\,T$ is $Tv$ with $v=\sum a_kv_k$, so $w=\sum a_kTv_k$. This is why the columns of a matrix span the range of the map it represents.

If $Tv=0$ then $T(Tv)=0$. If $w=Tv$, then $Tw\in\mathrm{range}\,T$. A line $\mathrm{span}(v)$ is invariant iff $Tv=\lambda v$ for some $\lambda$.

$T^kv=\lambda^kv$ by induction, then take linear combinations. Over $\mathbf C$ the converse holds too: every eigenvalue of $p(T)$ is $p(\lambda)$ for some eigenvalue $\lambda$ of $T$.

Compute $T^2$.

$T^2(x,y)=T(-3y,x)=(-3x,-3y)$, so $T^2=-3I$. If $Tv=\lambda v$ with $v\ne0$, then $-3v=T^2v=\lambda^2v$, so $\lambda^2=-3$, impossible for real $\lambda$. So $T$ has no eigenvalues. (Regarded on $\mathbf C^2$, the eigenvalues are $\pm i\sqrt3$.)

Try monomials.

$Tx^k=x\cdot kx^{k-1}=kx^k$. So $0,1,2,3,4$ are eigenvalues, with eigenvectors $x^k$. These are five distinct eigenvalues on a 5-dimensional space, so there are no others (Axler 5.12), each eigenspace is 1-dimensional, and the eigenvectors are exactly the nonzero multiples $cx^k$.

The minimal polynomial divides any polynomial that kills $T$. $T$ is diagonalizable iff its minimal polynomial has distinct roots (Axler 5.62).

(a) The minimal polynomial divides $z^4-1=(z-1)(z+1)(z-i)(z+i)$, which has distinct roots, so so does the minimal polynomial.

(b) It divides $z^4-z=z(z-1)(z-\omega)(z-\bar\omega)$ with $\omega=e^{2\pi i/3}$, again distinct roots.

(c) $T=\begin{pmatrix}0&1\\0&0\end{pmatrix}$: $T^2=0$, so $T^4=T^2$. Its minimal polynomial is $z^2$, a repeated root; its only eigenvector direction is $e_1$.

Expand both squares; the cross terms cancel. For $\ell^1$, take $u=e_1$, $v=e_2$: $\|u+v\|^2+\|u-v\|^2=4+4=8\ne2+2$.

$v=P_Uv+(v-P_Uv)$ with the second term in $U^\perp$; and $U\cap U^\perp=0$ since $\langle u,u\rangle=0\Rightarrow u=0$. Then dimensions add (Axler 3.94).

First take a single nonzero coefficient, then two.

Taking $a_j=1$ and the rest 0 gives $\|e_j\|=1$. For $j\ne k$ and a scalar $t$, take $a_j=1$, $a_k=t$: $\|e_j+te_k\|^2=1+|t|^2+2\,\mathrm{Re}\big(\bar t\langle e_j,e_k\rangle\big)$, and the hypothesis says this is $1+|t|^2$. So $\mathrm{Re}(\bar t\langle e_j,e_k\rangle)=0$ for every $t$. With $t=1$ the real part of $\langle e_j,e_k\rangle$ vanishes, and with $t=i$ (in the complex case) the imaginary part does. So $\langle e_j,e_k\rangle=0$.

The condition determines $e_k$ up to sign: it must be a unit vector in a 1-dimensional subspace.

Let $U_k=\mathrm{span}(v_1,\dots,v_k)$. If the list works, then $e_k\in U_k$ and $e_k\perp U_{k-1}$ (since $U_{k-1}=\mathrm{span}(e_1,\dots,e_{k-1})$). Now $U_k\cap U_{k-1}^\perp$ is 1-dimensional (Axler 6.51 inside $U_k$), so $e_k$ is one of its two unit vectors. That gives at most $2^m$ lists.

Conversely, Gram–Schmidt gives one valid list $e_1,\dots,e_m$, and changing any signs keeps it orthonormal with the same spans. So all $2^m$ sign choices work, and there are exactly $2^m$.

One inclusion is immediate since each $v_k$ is in the span. For the other, use linearity of the inner product.

Being orthogonal to the larger set $\mathrm{span}(v_k)$ implies being orthogonal to each $v_k$, so the right side is inside the left. If $u\perp v_k$ for every $k$, then $\langle u,\sum a_kv_k\rangle=\sum\bar a_k\langle u,v_k\rangle=0$, so the left side is inside the right.

$\lambda\|v\|^2=\langle Tv,v\rangle=\langle v,Tv\rangle=\bar\lambda\|v\|^2$. If $Tv=\lambda v$, $Tw=\mu w$ with $\lambda\ne\mu$ real: $\lambda\langle v,w\rangle=\langle Tv,w\rangle=\langle v,Tw\rangle=\mu\langle v,w\rangle$, so $\langle v,w\rangle=0$.

With $v=\sum a_ke_k$ (extended to an orthonormal basis), $\|Tv\|^2=\sum s_k^2|a_k|^2\le s_1^2\sum|a_k|^2$, with equality at $v=e_1$.

Use $\langle Tu,w\rangle=\langle u,T^*w\rangle$, and for the converse apply the first half to $T^*$.

Suppose $TU\subseteq U$. For $w\in U^\perp$ and $u\in U$: $\langle u,T^*w\rangle=\langle Tu,w\rangle=0$, since $Tu\in U$. So $T^*w\in U^\perp$.

Conversely, if $U^\perp$ is invariant under $T^*$, the first half applied to $T^*$ and $U^\perp$ says $(U^\perp)^\perp=U$ is invariant under $T^{**}=T$ (Axler 6.52, 7.5).

The singular values are the square roots of the eigenvalues of $T^*T$, and $\|Tv\|^2=\langle T^*Tv,v\rangle$.

If $T=0$, then $T^*T=0$, whose eigenvalues are all 0. Conversely, if all singular values are 0, the self-adjoint operator $T^*T$ has only the eigenvalue 0. By the spectral theorem it is diagonal with zeros in an orthonormal basis, so $T^*T=0$. Then $\|Tv\|^2=\langle T^*Tv,v\rangle=0$ for every $v$, so $T=0$.

A nilpotent map with a large entry.

$T=\begin{pmatrix}0&5\\0&0\end{pmatrix}$, i.e. $T(z_1,z_2)=(5z_2,0)$. It is upper triangular with zero diagonal, so 0 is its only eigenvalue. $T^*T=\begin{pmatrix}0&0\\0&25\end{pmatrix}$, so the singular values are 5 and 0. Eigenvalues and singular values agree for normal operators, but this $T$ is far from normal.

$T^2=\begin{pmatrix}a^2+bc&ab+bd\\ca+dc&cb+d^2\end{pmatrix}$. Subtract $(a+d)T$ and add $(ad-bc)I$: every entry becomes 0, e.g. $a^2+bc-a^2-ad+ad-bc=0$.