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The number of linearly independent eigenvectors for eigenvalue $1$ for the given matrix ?

$\begin{bmatrix} 1 & 3 & 2 \\ 0 & 4 & 2\\ 0 &-3 & -1 \end{bmatrix}$

Eigenvalues are $1,1,2$ for the above matrix .So, $1$ has multiplicity $2$ and $2$ has multiplicity $1$.

After putting the value $1$, I am getting $ i= $any value, and the relation $3j = -2k$.

Now, How to proceed further ?

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    **Hint:** You **do not** have a deficient matrix here and can find three linearly independent eigenvectors, even given the repeated eigenvalue. $\lambda_{1,2,3} = 1,1,2$, with eigenvectors $v_1 = (1, 0, 0), v_2 = (0,-2,3), v_3 = (-1,-1,1)$. – 2017-02-05
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    @copper.hat Can you explain ? – 2017-02-05
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    @JonGarrick: I can't, I made a mistake in computation. – 2017-02-05
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    @JonGarrick: As Moo pointed out, the are two li. eigenvectors of $A$ for the eigenvalue 1. Look at $\ker (A-I)$, it has dimension 2. – 2017-02-05
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    @copper.hat One more thing, Is this a rule of thumb to find rank of $A−λI$ for Li. inde. EV ? – 2017-02-05
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    @JonGarrick: I'm not sure what you are asking, $\operatorname{dim}\ker(A-\lambda I)$ will be the number of linearly independent eigenvectors. – 2017-02-05
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    @copper.hat Thanks :) Just wanted that .. – 2017-02-05

1 Answers 1

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Find the rank of $A-I$:

$$\begin{pmatrix} 0 && 3 && 2 \\ 0 && 3 && 2 \\ 0 && - 3 && -2 \end{pmatrix} $$

which is clearly $1=3-2$, so the size of the eigenspace is $2$-dimensional

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    Why take $A -I$ ? – 2017-02-05
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    @JonGarrick because the eigenvalue in question is $1$. If it were $2$ I would take $A-2I$ and if it were $\lambda$ I would look at $A-\lambda I$. – 2017-02-05
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    Okk !! Got now. One more thing, Is this a rule of thumb to find rank of $A−λI$ ? – 2017-02-05
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    @JonGarrick indeed this is more than that, it's essentially the definition of the co-dimension of the eigenspace for $\lambda$ (the "co" means you have to take the total dimension minus the rank, i.e. $3-1=2$). – 2017-02-05
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    Well, in general, find the rank of $A-\lambda I$. – 2017-02-05
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    @AdamHughes Thanks :-) Learn't something new today :) – 2017-02-05