Best solutions to - "Find a 3×3 matrix A that has eigenvalues..."

Falak Kinney

Answered question

2021-03-02

Find a $3 \times 3$ matrix A that has eigenvalues $λ = 0$ , 4 ,-4 with corresponding eigenvectors
$[\begin{matrix} 0 \\ 1 \\ - 1 \end{matrix}], [\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix}], [\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}]$

Answer & Explanation

crocolylec

Skilled2021-03-03Added 100 answers

Step 1
We need to find a matrix A which has eigenvalues $λ = 0, 4, - 4$ and eigenvectors as
$[\begin{matrix} 0 \\ 1 \\ - 1 \end{matrix}], [\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix}], [\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}]$
Now, a matrix is similar to a diagonal matrix D if there exist matrices P and $P^{- 1} s u c h t h a t A = P D P^{- 1}$
Here, diagonal matrix D consists of eigenvalues as the diagonal entries.
And the matrix P is the matrix that has eigenvectors as its columns.
Hence,
$D = [\begin{matrix} 0 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & - 4 \end{matrix}] and$
$P = [\begin{matrix} 0 & 1 & 0 \\ 1 & - 1 & 1 \\ - 1 & 1 & 1 \end{matrix}]$
Step 2
Hence, the matrix A is
$A = P D P^{- 1}$
$= [\begin{matrix} 0 & 1 & 0 \\ 1 & - 1 & 1 \\ - 1 & 1 & 1 \end{matrix}] [\begin{matrix} 0 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & - 4 \end{matrix}] {[\begin{matrix} 0 & 1 & 0 \\ 1 & - 1 & 1 \\ - 1 & 1 & 1 \end{matrix}]}^{- 1}$
Before finding A we need to find $P^{- 1} = {[\begin{matrix} 0 & 1 & 0 \\ 1 & - 1 & 1 \\ - 1 & 1 & 1 \end{matrix}]}^{- 1}$
Now, inverse of a matrix P is $P^{- 1} = \frac{a d j (P)}{| P |}$
Now,
$a d j (P) = {[\begin{matrix} + (- 1 \times 1 - 1 \times 1) & - (1 \times 1 - 1 \times (- 1)) & + (1 \times 1 - (- 1) \times (- 1)) \\ - (1 \times 1 - 0 \times (- 1)) & + (0 \times 1 - 0 \times (- 1)) & - (0 \times 1 - 1 \times (- 1)) \\ + (1 \times 1 - 0 \times (- 1)) & - (0 \times 1 - 0 \times 1) & + (0 \times (- 1) - 1 \times 1) \end{matrix}]}^{T}$
$= {[\begin{matrix} - 2 & - 2 & 0 \\ - 1 & 0 & - 1 \\ 1 & 0 & - 1 \end{matrix}]}^{T}$
$= [\begin{matrix} - 2 & - 1 & 1 \\ - 2 & 0 & 0 \\ 0 & - 1 & - 1 \end{matrix}]$
And
$| P | = | \begin{matrix} 0 & 1 & 0 \\ 1 & - 1 & 1 \\ - 1 & 1 & 1 \end{matrix} |$
$= 0 | \begin{matrix} - 1 & 1 \\ 1 & 1 \end{matrix} | - | \begin{matrix} 1 & 1 \\ - 1 & 1 \end{matrix} | + 0 | \begin{matrix} 1 & - 1 \\ - 1 & 1 \end{matrix} | = - 2$
Hence,

Jeffrey Jordon

Expert2022-01-30Added 2605 answers

Answer is given below (on video)

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