Question

1. Find a 3 × 3 matrix A that has eigenvalues λ = 0, 6, − 6 with corresponding eigenvectors

0 0 0
1 , -1 , 1
-1 1 1

2. Let Upper T ⁢ colon Upper M Subscript 22 Baseline ⁢ right-arrow Upper M Subscript 22 be the dilation operator with factor k = 3.

(a) Find Upper T left-parenthesis Start 2 By 2 Matrix 1st Row 1st Column 1, 2nd Column 4 2nd Row 1st Column -7, 2nd Column 4 End Matrix right-parenthesis = ????

(b) Find the rank and nullity of T .

3. Let Upper T ⁢ colon Upper P Subscript 2 Baseline ⁢ right-arrow Upper P Subscript 2 be the contraction operator with factor k ⁢ equals StartFraction 1 Over 6 EndFraction Number .

(a) Find Upper T left-parenthesis 1 ⁢ plus 6 x ⁢ plus 12 x Superscript 2 Baseline right-parenthesis Number .

(b) Find the rank and nullity of T.

Answer 1

A diagonalizable matrix is diagonalized by a matrix of its eigenvectors. So $A=P\Lambda P^{-1 }$ where $P$ is the matrix whose columns are the eigenvectors of $A$ and $\Lambda$ is a diagonal matrix whose diagonal entries are the eigenvalues of $A.$

$P=\begin{pmatrix} 1 & 0 & 0 \\ 1 & -1 & 1 \\ -1 & 1 & 1 \end{pmatrix} , \Lambda=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & -6 \end{pmatrix}$

Find $P^{-1}$

$$\begin{vmatrix} 1 & 0 & 0 \\ 1 & -1 & 1 \\ -1 & 1 & 1 \end{vmatrix}=1\begin{vmatrix} -1 & 1 \\ 1 & 1 \end{vmatrix}=1(-1-1)=-2\not=0$$

$C_{11}=(-1)^{1+1}\begin{vmatrix} -1 & 1 \\ 1 & 1 \end{vmatrix}=-2,$

$C_{12}=(-1)^{1+2}\begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix}=-2,$

$C_{13}=(-1)^{1+3}\begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix}=0,$

$C_{21}=(-1)^{2+1}\begin{vmatrix} 0 & 0 \\ 1 & 1 \end{vmatrix}=0,$

$C_{22}=(-1)^{2+2}\begin{vmatrix} 1 & 0 \\ -1 & 1 \end{vmatrix}=1,$

$C_{23}=(-1)^{2+3}\begin{vmatrix} 1 & 0 \\ -1 & 1 \end{vmatrix}=-1,$

$C_{31}=(-1)^{3+1}\begin{vmatrix} 0 & 0 \\ -1 & 1 \end{vmatrix}=0,$

$C_{32}=(-1)^{3+2}\begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix}=-1,$

$C_{33}=(-1)^{3+3}\begin{vmatrix} 1 & 0 \\ 1 & -1 \end{vmatrix}=-1.$

$$C=\begin{pmatrix} -2 & -2 & 0 \\ 0 & 1 & -1 \\ 0 & -1 & -1 \end{pmatrix}$$

The adjount matrix is:

$$\begin{pmatrix} -2 & 0& 0 \\ -2 & 1 & -1 \\ 0 & -1 & -1 \end{pmatrix}$$ $$P^{-1}=\begin{pmatrix} 1 & 0& 0 \\ 1 & -0.5 & 0.5 \\ 0 & 0.5 & 0.5 \end{pmatrix}$$

$P\Lambda=\begin{pmatrix} 1 & 0 & 0 \\ 1 & -1 & 1 \\ -1 & 1 & 1 \end{pmatrix}\begin{pmatrix} 0 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & -6 \end{pmatrix}=$

$=\begin{pmatrix} 0 & 0 & 0 \\ 0 & -6 & -6 \\ 0 & 6 & -6 \end{pmatrix}$

$A=P\Lambda P^{-1 }=\begin{pmatrix} 0 & 0 & 0 \\ 0 & -6 & -6 \\ 0 & 6 & -6 \end{pmatrix} \begin{pmatrix} 1 & 0& 0 \\ 1 & -0.5 & 0.5 \\ 0 & 0.5 & 0.5 \end{pmatrix}=$

$=\begin{pmatrix} 0 & 0 & 0 \\ -6 & 0 & -6 \\ 6 & -6 & 0 \end{pmatrix}$

$$A=\begin{pmatrix} 0 & 0 & 0 \\ -6 & 0 & -6 \\ 6 & -6 & 0 \end{pmatrix}$$

2.

(a) Let $T:M_{22}\to M_{22}$ be the dilation operator with factor k = 3.

This means that $T(A)=3A \ \ \ \forall A\in M_{22}$

$$A=\begin{pmatrix} 1 & 4 \\ -7 & 2 \end{pmatrix}$$

Then

$$T(A)=3A=3\begin{pmatrix} 1 & 4 \\ -7 & 2 \end{pmatrix}=\begin{pmatrix} 3 & 12 \\ -21 & 6 \end{pmatrix}$$

$$T(A)=\begin{pmatrix} 3 & 12 \\ -21 & 6 \end{pmatrix}$$

(b) Find the rank and nullity of T.

As we know, nullity represents dimension of Kernel.

We have :

$$B\in Ker{\{T\}}=>T(B)=0$$ $$=>3B=0=>Ker{\{T\}}=0$$

Which means that

$$nullity(T)=0$$

$$dim(M_{22})=rank(T)+nullity(T)$$ $$dim(M_{22})=4, nullity(T)=0$$

Hence

$$rank(T)=dim(M_{22})-nullity(T)=4-0=4$$ $$rank(T)=4$$

3.

Let $T:P_{2}\to P_{2}$ be the contraction operator with factor k = 1/6.

Find $T(1+6x+12x^2)$

$$T(p(x))={1 \over 6}p(x),\ \forall p(x)\in P_2$$

That gives us:

$$T(1+6x+12x^2)={1 \over 6}(1+6x+12x^2)={1 \over 6}+x+2x^2$$ $$T(1+6x+12x^2)={1 \over 6}+x+2x^2$$

(b) Find the rank and nullity of T.

As we know, nullity represents dimension of Kernel.

We have :

$$p\in Ker{\{T\}}=>T(p(x))=0, \forall x$$ $$=>{1 \over 6}p(x)=0=>Ker{\{T\}}=0$$

Which means that

$$nullity(T)=0$$

$$dim(P_{2})=rank(T)+nullity(T)$$ $$dim(P_{2})=3, nullity(T)=0$$

Hence

$$rank(T)=dim(P_{2})-nullity(T)=3-0=3$$ $$rank(T)=3$$