Determine whether A is similar B .If A sim B, give an invertible matrix P such that P^{-1}AP = B A=begin{bmatrix}1 & 1&0 0&1&10&0&1 end{bmatrix} , B=begin{bmatrix}1&1&0 0&1&00&0&1 end{bmatrix}

Determine whether A is similar B .If A sim B, give an invertible matrix P such that P^{-1}AP = B A=begin{bmatrix}1 & 1&0 0&1&10&0&1 end{bmatrix} , B=begin{bmatrix}1&1&0 0&1&00&0&1 end{bmatrix}

Question
Matrices
asked 2020-10-26
Determine whether A is similar B .If \(A \sim B\), give an invertible matrix P such that \(P^{-1}AP = B\)
\(A=\begin{bmatrix}1 & 1&0 \\0&1&1\\0&0&1 \end{bmatrix} , B=\begin{bmatrix}1&1&0 \\0&1&0\\0&0&1 \end{bmatrix}\)

Answers (1)

2020-10-27
Step 1
The given matrices are \(A=\begin{bmatrix}1 & 1&0 \\0&1&1\\0&0&1 \end{bmatrix} \text{ and } B=\begin{bmatrix}1&1&0 \\0&1&0\\0&0&1 \end{bmatrix}\)
Here, it is observed that the matrices A and B are diagonal matrix.
Therefore, the eigenvalues are the entries on the main diagonal values.
Hence, the eigenvalues of both A, B are \(\lambda_1=1,\lambda_2=1 and \lambda_3=1\)
Step 2
Obtain the eigenvector corresponding to the eigenvalue \(\lambda=1\) by solving the system (A-I)X=0. Let \(X=\begin{bmatrix}x_1 \\x_2 \\x_3 \end{bmatrix}\)
Then, \(\left(\begin{bmatrix}1 & 1&0 \\0&1&1\\0&0&1 \end{bmatrix}-\begin{bmatrix}1&1&0 \\0&1&0\\0&0&1 \end{bmatrix}\right)\begin{bmatrix}x_1 \\x_2 \\x_3 \end{bmatrix}=0\)
\(\begin{bmatrix}0&1&0 \\0&0&1\\0&0&0 \end{bmatrix}\begin{bmatrix}x_1 \\x_2 \\x_3 \end{bmatrix}=0\)
The system of equations corresponding to the above system is \(x_1=x_1\)(free variable)
\(x_2=0\)
\(x_3=0\)
If \(x_1=1 \text{ then } v_1= \begin{bmatrix}1 \\0 \\0 \end{bmatrix}\)
Step 3
Obtain the eigenvector corresponding to the eigenvalue \(\lambda=1\) by solving the system (B-I)X=0.
Then \(\left(\begin{bmatrix}1 & 1&0 \\0&1&0\\0&0&1 \end{bmatrix}-\begin{bmatrix}1&0&0 \\0&1&0\\0&0&1 \end{bmatrix}\right)\begin{bmatrix}x_1 \\x_2 \\x_3 \end{bmatrix}=0\)
\(\begin{bmatrix}0&1&0 \\0&0&0\\0&0&0 \end{bmatrix}\begin{bmatrix}x_1 \\x_2 \\x_3 \end{bmatrix}=0\)
The system of equations corresponding to the above system is \(x_1=x_1\)(free variable)
\(x_2=0\)
\(x_3=x_3\)(free variable)
If \(x_1=1 , x_3=0 \text{ then } v_1= \begin{bmatrix}1 \\0 \\0 \end{bmatrix}\)
Choose the arbitrary value \(x_1=0 , x_3=1\)
Then the eigenvector is \(v_2= \begin{bmatrix}0 \\0 \\1 \end{bmatrix}\)
It is observed that the eigenspace corresponding to \(\lambda=1\) of a matrix B has dimension 2 and the eigenspace corresponding to \(\lambda=1\) of a matrix A has dimension 1.
Hence, the matrices are not similar.
0

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