Question

Can someone confirm my work?

is this matrix diagonalizable?

Answer 1

given matrix is ( I didn't include brackets)

1 0 1
0 2 0
-1 0 -1

I did :
\[\lambda(I)-A = \left[\begin{matrix}\lambda-1 & 0 & 1 \\ 0 & \lambda-2 & 0\\ 1 & 0 & \lambda-1\end{matrix}\right]\]

\[\det(\lambda(I)-A) = \lambda-2(-1)^{2+2} \left[\begin{matrix}\lambda-1 & -1 \\ 1 & 1\end{matrix}\right]\]
\[= \lambda-2(1) \left| \lambda-1 - (1(-1)) \right|\]
\[= \lambda-2 (\lambda-1+1)\]
\[=\lambda-2(\lambda)\]
eigenvalues :
\[\lambda=0, \lambda=2\]

Answer 2

Then I solved for the eigenvectors by finding the basis:

When \[\lambda=0 \]
\[\left[\begin{matrix}-1 & 0 & -1 \\ 0 & -2 & 0 \\ 1 & 0 & -1 \end{matrix}\right]\]


wooops. I messed up! on the \[\lambda(I)-A \] part... forgot to do something with the third row

Answer 3

Yes bro I is I think you are orrect

Answer 4

Wait you done sompthing wrong

Answer 5

If I have \[\lambda ^{2}(\lambda-2)\]

would my eigenvalues be
\[\lambda _{1} = 0, \lambda _{2}= 0 , \lambda _{3}=2\] ?

Answer 6

solving for my eigenvectors:

when my lambda = 0


\[\left[\begin{matrix}-1 & 0 & -1 \\ 0 & -2 & 0\\ 1 & 0 & 1 \end{matrix}\right]\]
Then I added Row 1 to Row 3

giving me :
\[\left[\begin{matrix}-1 & 0 & -1 \\ 0 & -2 & 0 \\ 0 & 0 & 0 \end{matrix}\right]\]

My solutions are :

\[x _{1} = -t , x _{2} = 0, x _{3} = t\]

When t = 1, my basis is
\[W _{0} = \left\{ (-1,0,1) \right\}\]

so it is not diagonalizable because I don't have 3 independent eigenvectors?

Answer 7

First,
\[\lambda I - A = \begin{pmatrix} \lambda-1& 0 &-1 \\0 & \lambda-2 & 0\\1 & 0 & \lambda+1 \end{pmatrix}\]
So,
\[|\lambda I - A| = (\lambda - 2) (-1)^2 \bigl(
(\lambda^2 - 1) - (-1)\bigr) = (\lambda - 2) \lambda^2\]
Eigenvalues: 0, 0, 2.

The eigenspaces are indeed \(\langle (0,1,0)\rangle\) and \(\langle (1,0,-1) \rangle\).

Your conclusion is correct. `(can't find three linearly independent eigenvectors for this matrix)`.