Homework 8

1. Find the algebraic and geometric multiplicity of the eigenvalues of the following matrices. $$A = \begin{bmatrix} 1 & 1 & −1 & 1 & 2 \\ 0 & 2 & 1 & 1 & 3 \\ 0 & 0 & 3 & 0 & 1 \\ 0 & 0 & 0 & 4 & −1 \\ 0 & 0 & 0 & 0 & −1\end{bmatrix},\quad B = \begin{bmatrix}1 & 1 & 1 \\0 & 1 & 1 \\0 & 0 & 2\end{bmatrix}$$Are they diagonalizable? Explain.

For $A = \begin{bmatrix} 1 & 1 & −1 & 1 & 2 \\ 0 & 2 & 1 & 1 & 3 \\ 0 & 0 & 3 & 0 & 1 \\ 0 & 0 & 0 & 4 & −1 \\ 0 & 0 & 0 & 0 & −1\end{bmatrix}$

$$A-\lambda I = \begin{bmatrix} 1-\lambda & 1 & −1 & 1 & 2 \\ 0 & 2-\lambda & 1 & 1 & 3 \\ 0 & 0 & 3-\lambda & 0 & 1 \\ 0 & 0 & 0 & 4-\lambda & −1 \\ 0 & 0 & 0 & 0 & −1-\lambda \end{bmatrix} \\ \det(A-\lambda I) = (1-\lambda)(2-\lambda)(3-\lambda)(4-\lambda)(−1-\lambda) = 0 \\ \therefore\lambda= -1, 1, 2, 3, 4$$

By observation, we can see that there are no repeated roots in the characteristic polynomial. As such, the algebraic multiplicity of all five eigenvalues are $1$.

For $\lambda_1=-1$,

$$\operatorname{rref} (A-(-1) I) = \begin{bmatrix} 1 & 0 & 0 & 0 & \frac{11}{15} \\[.5em] 0 & 1 & 0 & 0 & \frac{59}{60} \\[.5em] 0 & 0 & 1 & 0 & \frac{1}{4} \\[.5em] 0 & 0 & 0 & 1 & -\frac{1}{5} \\[.5em] 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$

There’s one free variable, so the geometric multiplicity for $\lambda_1 = -1$ is $1$ (because the nullity i.e., $\dim\ker (A-\lambda)$ is $1$).

And similarly for $\lambda_2,\lambda_3,\lambda_4,\lambda_5$, we find that they all have one free variable.

$$\begin{array}{c} \lambda_2 = 1 & \lambda_3 = 2 & \lambda_4 = 3 & \lambda_5 = 4 & \\ \operatorname{rref}(A-I) & \operatorname{rref}(A-2I) & \operatorname{rref}(A-3I) & \operatorname{rref}(A-4I) \\ \\ \hline \\ \begin{bmatrix} 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} & \begin{bmatrix} 1 & -1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} & \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} & \begin{bmatrix} 1 & 0 & 0 & -\frac{1}{2} & 0 \\[.2em] 0 & 1 & 0 & -\frac{1}{2} & 0 \\[.2em] 0 & 0 & 1 & 0 & 0 \\[.2em] 0 & 0 & 0 & 0 & 1 \\[.2em] 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{array}$$

As such, the geometric multiplicity of all five eigenvalues are also $1$.

Since the algebraic and geometric multiplicity are the same for all five eigenvalues (all one), $A$ is diagonalizable.

For $B = \begin{bmatrix}1 & 1 & 1 \\0 & 1 & 1 \\0 & 0 & 2\end{bmatrix}$

$$B-\lambda I = \begin{bmatrix} 1-\lambda & 1 & 1 \\ 0 & 1-\lambda & 1 \\ 0 & 0 & 2-\lambda \end{bmatrix} \\ \begin{align*} \det(B-\lambda I) &= (1-\lambda)(1-\lambda)(2-\lambda) &= 0 \\ &= (1-\lambda)^2(2-\lambda) &= 0 \end{align*} \\ \therefore\lambda=1,2$$

Here, we see that $1$ is a repeated root. So, the algebraic multiplicity for:

• $\lambda_1=1$ is $2$ and
• $\lambda_2=2$ is $1$.

For $\lambda_1=1$,

$$\operatorname{rref}(B-I)= \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$

There’s one free variable, so the geometric multiplicity for $\lambda_1 = 1$ is $1$.

Similarly for $\lambda_2=2$,

$$\operatorname{rref}(B-2I) = \begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}$$

There’s one free variable, so the geometric multiplicity for $\lambda_2 = 2$ is $1$.

Since the algebraic and geometric multiplicity for $\lambda_1=1$ are not equal ($2$ and $1$, respectively), $B$ is not diagonalizable.

2. Find the conditions on $a$, $b$, $c$ so that the following matrix is diagonalizable.$$\begin{bmatrix} 1 & a & b \\ 0 & 2 & c \\ 0 & 0 & 1\end{bmatrix}$$

Let $A = \begin{bmatrix} 1 & a & b \\ 0 & 2 & c \\ 0 & 0 & 1 \end{bmatrix}$. $A$ is diagonalizable if the algebraic and geometric multiplicities for all eigenvalues are equal. By inspection, we have that:

$$\det (A-\lambda I) = (1-\lambda)^2 (2-\lambda) = 0 \\ \therefore\lambda = 1,2$$

So the eigenvalues are $1$ and $2$, with algebraic multiplicities of two and one, respectively.

For $\lambda_1 = 1$,

$$A-I = \begin{bmatrix} 0 & a & b \\ 0 & 1 & c \\ 0 & 0 & 0 \end{bmatrix} \implies \operatorname{rref}(A-I) = \begin{bmatrix} 0 & 1 & c \\ 0 & 0 & b-ac \\ 0 & 0 & 0 \end{bmatrix}$$

$A$ is diagonalizable if there exists two free variables for $\lambda_1=1$. The last column will be a non-pivot if $b-ac=0$.

For $\lambda_2=2$,

$$A-2I = \begin{bmatrix} -1 & a & b \\ 0 & 0 & c \\ 0 & 0 & -1 \end{bmatrix} \implies\operatorname{rref}(A-2I) = \begin{bmatrix} 1 & -a & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$

Additionally, for $\lambda_2=2$, there must be only one free variable. Luckily, the rank is already $2$ and $-a$ is already in a non-pivot position, so there’s no further restriction on $a$.

As such, the matrix is diagonalizable if $b-ac=0$.

3. Consider the set of all $3\times3$ upper triangular matrices

$$\mathcal{U} = \Set{\begin{bmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & f \end{bmatrix} : a, b, c, d, e, f \in \R }.$$

(a) Show that $\mathcal{U}$ is a subspace of $\mathcal{M}_{3\times3}$.

Checking if $\mathcal{U}$ is closed under addition

Consider two matrices $A,B\in\mathcal{U}$:

$$A = \begin{bmatrix} a_1 & b_1 & c_1 \\ 0 & d_1 & e_1 \\ 0 & 0 & f_1 \end{bmatrix}, \quad B = \begin{bmatrix} a_2 & b_2 & c_2 \\ 0 & d_2 & e_2 \\ 0 & 0 & f_2 \end{bmatrix}$$

Sure enough, adding $A$ and $B$ will still produce an upper-triangular matrix.

$$A + B = \begin{bmatrix} a_1 & b_1 & c_1 \\ 0 & d_1 & e_1 \\ 0 & 0 & f_1 \end{bmatrix} + \begin{bmatrix} a_2 & b_2 & c_2 \\ 0 & d_2 & e_2 \\ 0 & 0 & f_2 \end{bmatrix} = \begin{bmatrix} a_1+a_2 & b_1+b_2 & c_1+c_2 \\ 0 & d_1+d_2 & e_1+e_2 \\ 0 & 0 & f_1+f_2 \end{bmatrix}$$

Since $A+B\in\mathcal{U}$, it is closed under addition.

Checking if $\mathcal{U}$ is closed under scalar multiplication

Consider a scalar $\alpha\in\R$ and a matrix $a\in\mathcal{U}$:

$$A = \begin{bmatrix} a_1 & b_1 & c_1 \\ 0 & d_1 & e_1 \\ 0 & 0 & f_1 \end{bmatrix}$$

Multiplying a scalar to an upper-triangular matrix will still produce an upper-triangular matrix.

$$\alpha A = \begin{bmatrix} \alpha a_1 & \alpha b_1 & \alpha c_1 \\ 0 & \alpha d_1 & \alpha e_1 \\ 0 & 0 & \alpha f_1 \end{bmatrix}$$

Since $\alpha A\in\mathcal{U}$, it is closed under scalar multiplication.

Hence, $\mathcal{U}$ is a subspace of $\mathcal{M}_{3\times3}$.

(b) What is the dimension of $\mathcal{U}$?

For $a,b,c,d,e,f\in\R$, the matrix $\begin{bmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & f \end{bmatrix}$ can be decomposed as:

$$\begin{split} \begin{bmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & f \end{bmatrix} &= a \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} + b \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} + c \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ &\quad+ d \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} + e \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} + f \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \end{split}$$

Note that these six $3\times3$ matrices are linearly independent. By observation, we can see that

$$\vec{\mathbf{0}} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \iff a=b=c=d=e=f=0.$$

As such, by definition, it follows that these six vectors forms a basis of $\mathcal{U}$. Which subsequently means that $\dim\mathcal{U}=6$.

(c)

(i) Is it true that any 7 matrices taken from $\mathcal{U}$ must be linearly dependent? Explain

Yes. As shown in (b), its dimension is six. Choosing more than six means that at least one matrix must be the same or is a multiple of the other.

(ii) Is it true that any 6 matrices taken from $\mathcal{U}$ must be a basis for $\mathcal{U}$? Explain

No. If all six are identical or are multiple of each other, then they may not be a basis of $\mathcal{U}$.

4. Consider the set of all continuous functions on the interval $[a, b]$, denoted by $C([a, b])$. Show that the set of all functions with mean value zero, i.e. $$M = \Set{ f: \frac{1}{b-a}\int_a^b f(x)dx = 0}$$ is a subspace of $C([a, b])$.

Checking if $M$ is closed under addition

Consider two functions $f,g\in M$.

By linearity of integrals,

$$\int(f+g) = \int f + \int g = 0 + 0.$$

Since $f+g\in M$, it is closed under addition.

Checking if $M$ is closed under scalar multiplication

Similarly, for a scalar $\alpha\in\R$ and a function $f\in M$.

By linearity, we know that

$$\int \alpha f = \alpha\int f = \alpha (0).$$

As such, $M$ is closed under scalar multiplication.

Hence, $M$ is a subspace of $C([a, b])$.

5. Determine if the following sets of vectors linearly independent in their own vector space.

(i) $x^2 −3, 2 −x, (x−1)^2$ on $\mathcal{P}_2$.

For $a,b,c\in\R$, suppose that

$$a(x^2 −3) + b(2-x) + c(x−1)^2 = 0 \quad\forall x\in\R.$$

If $x=0$, then:

$$a(0^2 −3) + b(2-0) + c(0−1)^2 = 0 \\ -3a+2b+c = 0$$

If $x=1$, then:

$$a(1^2 −3) + b(2-1) + c(1−1)^2 = 0 \\ -2a+b = 0$$

If $x=2$, then:

$$a(2^2 −3) + b(2-2) + c(2−1)^2 = 0 \\ a+c = 0$$

And so, we have a homogenous system of equations:

$$\begin{cases} -3a+2b+c &= 0 \\ -2a+b &= 0 \\ a+c &= 0 \end{cases}$$

$$\operatorname{rref}\left[ \begin{array}{ccc|c} -3 & 2 & 1 & 0 \\ -2 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \end{array} \right] = \left[ \begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]$$

Since there are free variables, they are linearly dependent.

(ii) $\begin{bmatrix}2 & 1 \\3 & 2\end{bmatrix},\begin{bmatrix}1 & 2 \\0 & 3\end{bmatrix},\begin{bmatrix}1 & 5 \\2 & 0\end{bmatrix}$ on $\mathcal{M}_{2\times2}$

For $a,b,c\in\R$, suppose that:

$$a\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} + b\begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} + c\begin{bmatrix} 1 & 5 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \quad\forall a,b,c\in\R.$$

Then:
$$\begin{align*} a\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} + b\begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} + c\begin{bmatrix} 1 & 5 \\ 2 & 0 \end{bmatrix} &= \begin{bmatrix} 2a+b+c & a+2b+5c \\ 3a+2c & 2a+3b \end{bmatrix} \\ &= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \end{align*}$$

Putting it into a homogenous system, we get:

$$\operatorname{rref} \left[ \begin{array}{ccc|c} 2 & 1 & 1 & 0 \\ 1 & 2 & 5 & 0 \\ 3 & 0 & 2 & 0 \\ 2 & 3 & 0 & 0 \end{array} \right] = \left[ \begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]$$

Since there are no free variables, they are linearly independent.

(iii) $e^x, e^{3x}$ on $C([0, 1])$.

For $a,b\in\R$, suppose that:

$$ae^x + be^{3x} = 0\quad\forall a,b\in\R.$$

Let $x=0$, then:

$$ae^0 + be^{0} = 0 \\ a+b=0$$

Let $x=1$, then:

$$ae^1 + be^{3} = 0 \\ ae+be^3=0$$

Again, putting them into a system, we get:

$$\operatorname{rref}\left[ \begin{array}{cc|c} 1 & 1 & 0 \\ e & e^3 & 0 \end{array} \right] = \left[ \begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right]$$

As such, they are linearly independent.

6. Let $$M = \Set{ \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}: a_{11} + a_{22} + a_{33} = 0}.$$

(i) Show that $M$ is a subspace for $\mathcal{M}_{3\times3}$.

Checking if $M$ is closed under addition

Consider two matrices $A,B\in M$:

$$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix},\quad B = \begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{bmatrix}$$

Then,

$$A+B = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} + \begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{bmatrix} =\begin{bmatrix} a_{11}+b_{11} & a_{12}+b_{12} & a_{13}+b_{13} \\ a_{21}+b_{21} & a_{22}+b_{22} & a_{23}+b_{23} \\ a_{31}+b_{31} & a_{32}+b_{32} & a_{33}+b_{33} \end{bmatrix}.$$

And so the trace of $A+B$ is:

$$\begin{align*} (a_{11}+b_{11}) + (a_{22}+b_{22}) + (a_{33}+b_{33}) &= (a_{11} + a_{22} + a_{33}) + (b_{11} + b_{22} + b_{33}) \\ &= 0+0 \\ &= 0 \end{align*}$$

Therefore, $M$ is closed under addition.

Checking if $M$ is closed under scalar multiplication

Consider a scalar $\alpha\in\R$ and a matrix $A\in M$ where

$$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}.$$

Then,

$$\alpha A = \begin{bmatrix} \alpha a_{11} & \alpha a_{12} & \alpha a_{13} \\ \alpha a_{21} & \alpha a_{22} & \alpha a_{23} \\ \alpha a_{31} & \alpha a_{32} & \alpha a_{33} \end{bmatrix}.$$

And so,

$$\alpha a_{11}+\alpha a_{22}+\alpha a_{33} = \alpha(a_{11}+a_{22}+a_{33}) = \alpha(0) = 0.$$

Since $\alpha A\in M$, it is closed under scalar multiplication.

Hence, $M$ is a subspace of $\mathcal{M}_{3\times3}$.

(ii) Find a basis for $M$ and what is the dimension of $M$?

For $\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}\in M$. Since $a_{11}+a_{22}+a_{33}=0$, then $a_{11}=-a_{22}-a_{33}$.

And so,

$$\begin{align*} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} &= \begin{bmatrix} -a_{22}-a_{33} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \\ &\begin{split} &=a_{22} \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} + a_{33} \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} + a_{12} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} + a_{13} \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ &\quad+ a_{21} \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} + a_{23} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} + a_{31} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix} + a_{32} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} \end{split} \end{align*}$$

A basis of $M$ is

$$\Set{ \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} }$$

and its dimension is eight.