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I have given a matrix $$ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -3 & 3 \end{pmatrix}. $$ We have to find 3 linearly independent eigen vector. I have calculated the eigen values that is 1 with multiplicity 3 . I have found the eigen vector $ \begin{pmatrix}1 & 1 & 1 \end{pmatrix}^T$. But I can't find the other two linearly independent eigen vectors.

Any help is appreaciating.

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    $\begingroup$ Have you learned about algebraic and geometric multiplicity? $\endgroup$ – mathreadler Apr 1 '17 at 14:14
  • $\begingroup$ You are right. tha matrix does not have three linearly independent eigenvectors. $\endgroup$ – Emilio Novati Apr 1 '17 at 14:18
  • $\begingroup$ yes it is actually algebraic multiplicity . I am searching about the other two generalised eigen vector $\endgroup$ – M. A. SARKAR Apr 1 '17 at 14:19
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The characteristic polynomial is \begin{align} \det\begin{pmatrix} 0-X & 1 & 0 \\ 0 & 0-X & 1 \\ 1 & -3 & 3-X \end{pmatrix} &= -X\det\begin{pmatrix}-X & 1 \\ -3 & 3-X\end{pmatrix} -\det\begin{pmatrix}0 & 1 \\ 1 & 3-X\end{pmatrix} \\ &=-X(-3X+X^2+3)+1 \\[6px] &=-X^3+3X^2-3X+1=(1-X)^3 \end{align} There cannot exist three linearly independent vectors: the matrix would be diagonalizable and similar to $$ \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} $$ but the identity matrix is only similar to itself.

Actually, if we compute the rank of $A-I$ (where $A$ is the given matrix), we get, with Gaussian elimination, $$ A-I=\begin{pmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 1 & -3 & 2 \end{pmatrix} \to \begin{pmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 0 & -2 & 2 \end{pmatrix} \to \begin{pmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 0 \end{pmatrix} $$ so the rank is $2$ and the eigenspace relative to $1$ has dimension $3-2=1$. No set of two eigenvalues can be linearly independent.

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  • $\begingroup$ ok thank you very much $\endgroup$ – M. A. SARKAR Apr 1 '17 at 14:42
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Furthermore the generalized eigenspace can be found through:

$$(A-I_3)^{-1}[1,1,1]^T = [-1,0,1]^T$$

Which is then the vector which has image split onto self and first eigenvector.

Iterating:

$$(A-I_3)^{-1}([-1,0,1]^T+[1,1,1]^T) = [-1,-1,2]^T$$

Now we have our $\bf T$, or as in Jordan normal form often called $\bf S$:

$${\bf A = SJS}^{-1}$$

$${\bf J} = \left[\begin{array}{ccc}\color{red} 1&\color{blue}1&0\\0&\color{red} 1&\color{blue}1\\0&0&\color{red} 1\end{array}\right]\hspace{1cm}{\bf S} = \left[\begin{array}{ccc} 1&0&- \frac 1 3\\ 1&1&-\frac 1 3\\ 1&2&\frac{2}3 \end{array}\right]$$

The red values in our Jordan block is the eigenvalue and the blue numbers are the off diagonal ones (those will always be one and have nothing to do with the eigenvalue).

EDIT as amd points ut the equation solved above $(A-I_3)^{-1}v$ actually is a pseudo inverse or least squares solution since $A-I_3$ is singular.

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  • $\begingroup$ That doesn’t look right. Since $1$ is an eigenvalue, $A-I_3$ is singular. $\endgroup$ – amd Apr 1 '17 at 20:55
  • $\begingroup$ You can calculate $SJS^{-1}$ it should be A unless my software is broken... $\endgroup$ – mathreadler Apr 1 '17 at 21:00
  • $\begingroup$ $(A-I_3)^{-1}$ doesn’t exist. I’m sure the Jordan decomposition that you ended up with is correct, though I haven’t bothered to check it. $\endgroup$ – amd Apr 1 '17 at 21:02
  • $\begingroup$ Yes you are right pseudoinerse or least squares solution is probably what my software did for me in place of that $(.)^{-1}$. $\endgroup$ – mathreadler Apr 1 '17 at 21:23

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