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Given

$$A = \begin{bmatrix} -9 & 4 & 4\\ -8 & 3 & 4 \\ -16 & 8 & 7 \end{bmatrix}$$

I calculated eigenvalues $\lambda= -1,-1,3$. The algebraic multiplicity (AM) and geometric multiplicity (GM) of $\lambda=-1$ are $2$, which tells us that there will be two linearly independent eigenvectors.

I am not sure how to find the eigenvectors. Usually, I take one variable and equate it to $t$ and then solve it for the other two. I am not quite sure how to find eigenvectors when we have two free variables.

Steps:

$$(A-\lambda I)=0$$

$$\begin{bmatrix} -8 & 4 & 4\\ -8 & 4 & 4 \\ -16 & 8 & 8 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$$

$$\begin{bmatrix} 8 & -4 & -4\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$$

$$2x-y-z=0$$

I don't know how to proceed from here.

Rodrigo de Azevedo
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Daman
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  • So, $2x-y-z=0$. Let us define some parametric variables. Out of convenience, lets use the parameters $s$ and $t$ and let us use $y=s$ and $z=t$. We have then $\begin{cases}x=\frac{1}{2}s+\frac{1}{2}t\y=1s+0t\z=0s+1t\end{cases}$. Can you continue now? – JMoravitz May 23 '18 at 04:15
  • Note: there are infinitely many correct answers. The answer I lead you to above is just one of many. You could have instead chosen to set $x=s$ and $y=t$ for example which would lead you to a different pair of eigenvectors. The point is that any correct answer will have the span of the two vectors all corresponding to the same eigenspace. – JMoravitz May 23 '18 at 04:18
  • Yes I understood little bit. I ll get back and ask if I am stuck again. @JMoravitz Thanks again . – Daman May 23 '18 at 04:36
  • @JMoravitz I have a problem. I am stuck. Three eigenvectors came $(1,1,0),(1,0,-1),(1,1,2)$. I have to Diagonalize it. My question is will it change determinant value if I take second vector$ (-1,0,1)$ in my second column instead of $(1,0,-1)$. – Daman May 23 '18 at 05:59
  • Of course the two determinants won’t be equal. You’re multiplying one of the columns by $-1$. Why does that matter? – amd May 23 '18 at 08:48
  • @amd That was silly I know but it will not alter diagonalizaton. – Daman May 23 '18 at 08:52
  • Suppose you diagonalize $A$ in two different ways: $A=SDS^{-1}$ as well as $A=TET^{-1}$ where $D$ and $E$ are both diagonal matrices whose entries are the eigenvalues of $A$ and $S$ and $T$ are both matrices whose columns are the corresponding eigenvectors. It is possible for $\det(S)\neq \det(T)$, sure. However... $\det(SDS^{-1})=\det(A)=\det(TET^{-1})$ so in that sense it doesn't matter at all which diagonalization you use. – JMoravitz May 23 '18 at 15:32
  • Note that $\det(A)=\det(SDS^{-1})=\det(S)\det(D)\det(S^{-1})=\det(S)\cdot det(D)\cdot\frac{1}{\det(S)}=\det(D)$ is equal to the product of the eigenvalues of $A$. The eigenvectors and the value of $\det(S)$ don't come into that calculation at all. There are in fact infinitely many correct ways in which you can diagonalize a matrix. Suppose $A=SDS^{-1}$ was a diagonalization. So too is $A=(2S)D(2S)^{-1}=(3S)D(3S)^{-1}=\dots$ etc... Each of $(S),(2S),(3S),\dots$ obviously have different determinants. – JMoravitz May 23 '18 at 15:37
  • As for your eigenvectors that you found, you must have made a mistake. None of $(1,1,0),(1,0,-1),(1,1,2)$ correspond to eigenvalue $-1$. – JMoravitz May 23 '18 at 15:42
  • See this answer for a way to read a basis for the null space from the rref matrix. – amd May 23 '18 at 18:32

2 Answers2

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so $x=\lbrace\frac{y}{2}+\frac{z}{2} $

let be $y=1$, $z=0$, then

$v1=\begin{bmatrix}\frac {1}{2} \\ {1}\\ {0} \end{bmatrix}$

let be $y=0$, $z=1$, then $v2=\begin{bmatrix}\frac {1}{2} \\ {0}\\ {1} \end{bmatrix}$

later, you only substitute $\lambda (i)$ and $v1$,$v2$ in the equations $\ (A-\lambda I)v1=0$

and $\ (A-\lambda I)v2=0$

Jessi
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The null space of a matrix is the orthogonal complement of its row space. The latter is clearly spanned by $(8,-4,-4)^T$, so you’re looking for a linearly independent pair of vectors orthogonal to this one. These can be found by inspection: for any vector $(a,b,c)^T$, the vectors $(0,c,-b)^T$, $(-c,0,a)^T$ and $(b,-a,0)^T$ are all orthogonal to it, and if $(a,b,c)^T\ne0$, then at least two of those orthogonal vectors are also nonzero.

amd
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