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## Solution to Linear Algebra Hoffman & Kunze Chapter 6.4

#### Exercise 6.4.1

Let $T$ be the linear operator on $\mathbb R^2$, the matrix of which in the standard ordered basis is
$$A=\left[\begin{array}{cc}1&-1\\2&2\end{array}\right].$$(a) Prove that the only subspaces of $\mathbb R^2$ invariant under $T$ are $\mathbb R^2$ and the zero subspace.

(b) If $U$ is the linear operator on $\mathbb C^2$, the matrix of which in the standard ordered basis is $A$, show that $U$ has $1$-dimensional invariant subspaces.

Solution: (a) The characteristic polynomial equals $$\left|\begin{array}{cc}x-1&1\\-2&x-2\end{array}\right|=(x-1)(x-2)+2=x^2-3x+4.$$ This is a parabola opening upwards with vertex $(3/2,7/4)$, so it has no real roots. If $T$ had an invariant subspace it would have to be $1$-dimensional and $T$ would therefore have a characteristic value.

(b) Over $\mathbb C$ the characteristic polynomial factors into two linears. Therefore over $\mathbb C$, $T$ has two characteristic values and therefore has at least one characteristic vector. The subspace generated by a characteristic vector is a $1$-dimensional subspace.

#### Exercise 6.4.2

Let $W$ be an invariant subspace for $T$. Prove that the minimal polynomial for the restriction operator $T_W$ divides the minimal polynomial for $T$, without referring to matrices.

Solution: The minimum polynomial of $T_W$ divides any polynomial $f(t)$ where $f(T_W)=0$. If $f$ is the minimum polynomial for $T$ then $F(T)v=0$ $\forall$ $v\in V$. Therefore, $f(T)w=0$ $\forall$ $w\in W$. So $f(T_W)w=0$ $\forall$ $w\in W$ since by definition $f(T_W)w=f(T)w$ for $w\in W$. Therefore, $f(T_W)=0$. Therefore the minimum polynomial for $T_W$ divides $f$.

#### Exercise 6.4.3

Let $c$ be a characteristic value of $T$ and let $W$ be the space of characteristic vectors associated with the characteristic value $c$. What is the restriction operator $T_W$?

Solution: For $w\in W$ the transformation $T(w)=cw$. Thus $T_W$ is diagonalizable with single characteristic value $c$. In other words under which it is represented by the matrix
$$\left[\begin{array}{cccc}c&&&\\&c& \Large0&\\& \Large0&\ddots&\\&&&c\end{array}\right]$$where there are dim$(W)$ $c$'s on the diagonal.

#### Exercise 6.4.4

Let
$$A=\left[\begin{array}{ccc}0&1&0\\2&-2&2\\2&-3&2\end{array}\right].$$Is $A$ similar over the field of real numbers to a triangular matrix? If so, find such a triangular matrix.

Solution: We have$$A^2=\left[\begin{array}{ccc}2&-2&2\\0&0&0\\-2&2&-2\end{array}\right].$$And $A^3=0$. Thus the minimal polynomial $x^3$ and the only characteristic value is $0$. We now follow the constructive proof of Theorem 5. $W=\{0\}$, $\alpha_1$ a characteristic vector of $A$ is $\left[\begin{array}{c}1\\0\\-1\end{array}\right]$. We need $\alpha_2$ such that $A\alpha_2\in\{\alpha_1\}$. $\alpha_2=\left[\begin{array}{c}-1\\1\\2\end{array}\right]$ satisfies $A\alpha2=\alpha_1$. Now need $\alpha_3$ such that $A\alpha_3\in\{\alpha_1,\alpha_2\}$. $\alpha_3=\left[\begin{array}{c}0\\0\\1\end{array}\right]$ satisfies $A\alpha_3=2\alpha_1+2\alpha_2$. Thus with respect to the basis $\{\alpha_1,\alpha_2,\alpha_3\}$ the transformation corresponding to $A$ is $\left[\begin{array}{ccc}0&1&2\\0&0&2\\0&0&0\end{array}\right]$.

#### Exercise 6.4.5

Every matrix $A$ such that $A^2=A$ is similar to a diagonal matrix.

Solution: $A^2=A$ $\Rightarrow$ $A$ satisfies the polynomial $x^2-x=x(x-1)$. Therefore the minimum polynomial of $A$ is either $x$, $x-1$ or $x(x-1)$. In all three cases the minimum polynomial factors into distinct linears. Therefore, by Theorem 6 $A$ is diagonalizable.

#### Exercise 6.4.6

Let $T$ be a diagonalizable linear opeartor on the $n$-dimensional vector space $V$, and let $W$ be a subspace which is invariant under $T$. Prove that the restriction operator $T_W$ is diagonalizable.

Solution: By the lemma on page 80 the minimum polynomial for $T_W$ divides the minimum polynomial for $T$. Now $T$ diagonalizable implies (by Theorem 6) that the minimum polynomial for $T$ factors into distinct linears. Since the minimum polynomial for $T_W$ divides it, it must also factor into distinct linears. Thus by Theorem 6 again $T_W$ is diagonalizable.

#### Exercise 6.4.7

Let $T$ be a linear operator on a finite-dimensional vector space over the field of complex numbers. Prove that $T$ is diagonalizable if and only if $T$ is annihilated by some polynomial over $\mathbb C$ which has distinct roots.

Solution: If $T$ is diagonalizable then its minimum polynomial is a product of distinct linear factors, and the minimal polynomial annihilates $T$. This proves $\Rightarrow$". Now suppose $T$ is annihilated by a polynomial over $\mathbb C$ with distinct roots. Since the base field is $\mathbb C$ this polynomial factors completely into distinct linear factors. Since the minimum polynomial divides this polynomial the minimum polynomial factors completely into distinct linear factors. Thus by Theorem 6, $T$ is diagonalizable.

#### Exercise 6.4.8

Let $T$ be a linear operator on $V$. If every subspace of $V$ is invariant under $T$, then $T$ is a scalar multiple of the identity operator.

Solution: Let $\{\alpha_i\}$ be a basis. The subspace generated by $\alpha_i$ is invariant thus $T\alpha_i$ is a multiple of $\alpha_i$. Thus $\alpha_i$ is a characteristic vector since $T\alpha_i=c_i\alpha_i$ for some $c_i$. Suppose $\exists$ $i,j$ such that $c_i\not=c_j$. Then $$T(\alpha_i+\alpha_j)=T\alpha_i+T\alpha_j=c_i\alpha_i+c_j\alpha_j=c(\alpha_i+\alpha_j).$$ Since the subspace generated by $\{\alpha_i,\alpha_j\}$ is invariant under $T$. Thus $c_i=c$ and $c_j=c$ since coefficients of linear combinations of basis vectors are unique. Thus $T\alpha_i=c\alpha_i$ $\forall$ $i$. Thus $T$ is $c$ times the identity operator.

#### Exercise 6.4.9

Let $T$ be the indefinite inntegral operator
$$(Tf)(x)=\int_0^xf(t)dt$$on the space of continuous functions on the interval $[0,1]$. Is the space of polynomial functions invariant under $T$? Ths space of differentiable functions? The space of functions which vanish at $x=1/2$?

Solution: The integral from $0$ to $x$ of a polynomial is again a polynomial, so the space of polynomial functions is invariant under $T$. The integral from $0$ to $x$ of a differntiable function is differentiable, so the space of differentiable functions is invariant under $T$. Now let $f(x)=x-1/2$. Then $f$ vanishes at $1/2$ but $\int_0^xf(t)dt=\frac12x^2-\frac12x$ which does not vanish at $x=1/2$. So the space of functions which vanish at $x=1/2$ is not invariant under $T$.

#### Exercise 6.4.10

Let $A$ be a $3\times3$ matrix with real entries. Prove that, if $A$ is not similar over $\mathbb R$ to a triangular matrix, then $A$ is similar over $\mathbb C$ to a diagonal matrix.

Solution: If $A$ is not similar to a tirangular matrix then the minimum polynomial of $A$ must be of the form $(x-c)(x^2+ax+b)$ where $x^2+ax+b$ has no real roots. The roots of $x^2+ax+b$ are then two non-real complex conjugates $z$ and $\bar z$. Thus over $\mathbb C$ the minimum polynomial factors as $(x-c)(x-z)(x-\bar z)$. Since $c$ is real, $c$, $z$ and $\bar z$ constintute three distinct numbers. Thus by Theorem 6 $A$ is diagonalizable over $\mathbb C$.

#### Exercise 6.4.11

True or false? If the triangular matrix $A$ is similar to a diagonal matrix, then $A$ is already diagonal.

Solution: False. Let $A=\left[\begin{array}{cc}1&1\\0&0\end{array}\right]$. Then $A$ is triangular and not diagonal. The characteristic polynomial is $x(x-1)$ which has distinct roots, so the minimum polynomial is $x(x-1)$. Thus by Theorem 6, $A$ is diagonalizable.

#### Exercise 6.4.12

Let $T$ be a linear operator on a finite-dimensional vector space over an algebraically closed field $F$. Let $f$ be a polynomial over $F$. Prove that $c$ is a characteristic value of $f(T)$ if and only if $c=f(t)$, where $t$ is a characteristic value of $T$.

Solution: Since $F$ is algebraically closed, the corollary at the bottom of page 203 implies there's a basis under which $T$ is represented by a triangular matrix $A$. $A=[a_{ij}]$ where $a_{ij}=0$ if $i>j$ and the $a_{ii}$, $i=1,\dots,n$ are the characteristic values of $T$. Now $f(A)=[b_{ij}]$ where $b_{ij}=0$ if $i>j$ and $b_{ii}=f(a_{ii})$ for all $i=1,\dots,n$. Thus the characteristic values of $f(A)$ are exactly the $f(c)$'s where $c$ is a characteristic value of $A$. Since $f(A)$ is a matrix representative of $f(T)$ in the same basis, we conclude the same thing about the transformation $T$.

#### Exercise 6.4.13

Let $V$ be the space of $n\times n$ matrices over $F$. Let $A$ be a fixed $n\times n$ matrix over $F$. Let $T$ and $U$ be the linear operators on $V$ defined by
\begin{alignat*}{1}
T(B) &= AB\\
U(B) &= AB-BA
\end{alignat*}(a) True or false? If $A$ is a diagonalizable (over $F$), then $T$ is diagonalizable.

(b) True or false? If $A$ is diagonalizable, then $U$ is diagonalizable.

Solution: (a) True by Exercise 10 Section 6.3 page 198 since by Theorem 6 diagonalizability depends entirely on the minimum polynomial.

(b) True. Find a basis so that $A$ is diagonal
$$A=\left[\begin{array}{cccc}c_1&&&\\&c_2&\Large0&\\&\Large0&\ddots&\\&&&c_n\end{array}\right].$$Let $B=[b_{ij}]$. Then $U(B)=AB-BA$. The $n^2$ matrices $B_{ij}$ such that $b_{ij}\not=0$ and all other entries equal zero form a basis for $V$. For any $B_{ij}$, $U(B_{ij})=AB_{ij}-B_{ij}A=[d_{i'j'}]$ where $$d_{i'j'}=c_{i'}b_{i'j'}-c_{j'}b_{i'j'}=(c_{i'}-c_{j'})b_{i'j'}.$$ Thus $d_{i'j'}\not=0$ only when $i'=i$ and $j'=j$. Thus $U(B_{ij})=(c_i-c_j)B_{ij}$. So $c_i-c_j$ is a characteristic value and $B_{ij}$ is a characteristic vector for all $i,j$. Thus $V$ has a basis of characteristic vectors for $U$. Thus $U$ is diagonalizable.