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How do I find generalized eigenvector of [math] \begin{bmatrix}

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How do I find generalized eigenvector of
[math]
\begin{bmatrix}
1 & 0\\
1 & 0
\end{bmatrix}
[/math]
>>
>>7973213
pls halp
also, what is inverse of
\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}
>>
blease be bait
>>
>>7973213
nice I am learning what's eigenvector atm

my question is can you find diagonalized matrix with respect to some basis of your matrix?
>>
>>7973215
same matrix because

your matrix times your matrix is eye matrix which is the same
>>
>>7973261
Thank :^)
>>7973251
Characteristic equation is
[math]
(1-\lambda)(\lambda) = 0
[/math]
so one eigenvector is
[math]
Nul(\begin{bmatrix}
0 & 0 \\
1 & -1
\end{bmatrix})
[/math]
or
[math]
\begin{bmatrix}
1 \\
1
\end{bmatrix}
[/math]
and the obvious 0 eigenvector
[math]
\begin{bmatrix}
0 \\
1
\end{bmatrix}
[/math]

so our coordinate matrix is
[math]
\begin{bmatrix
1 & 0 \\
1 & 1 \\
[/end]
Thread replies: 6
Thread images: 1

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