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I have the matrix $$begin{pmatrix}3&-9\-9&27end{pmatrix}.$$ I found the eigenvalues of $0$ and $30.$ However, when I try to plug in $0$ for the Eigenvalues and row reduce, I get $0,0$ as my solutions for find $x_1$ and $x_2.$ This is not correct and when I tried to do it with the eigenvalue of $30,$ it also came out to $0,0.$ I know to find the unit you have to take the length of $0$ and $30.$ Any help would be appreciated.

In oerder to get an eigenvector whose eigenvalue is $0$, you solve the system$$left{begin{array}{l}3x-9y=0\-9x+27y=0end{array}right.$$Since the second equation is just the first one times $-3$, this is equivalent to having to deal only with the first equation. So, take $x=3$ and $y=1$, for instance. Problem: $(3,1)$ is not unitary. So, divide this vector by its norm, thereby getting $left(frac3{sqrt{10}},frac1{sqrt{10}}right)$.

The case of the other eigenvalue is similar.

Your answers quite make sense. According to the definition$$Av=lambda v$$where the notation is self-explaining. If $lambda=0$ for the matrix in this question we obtain$$begin{pmatrix}3&-9\-9&27end{pmatrix}begin{pmatrix}v_1\v_2end{pmatrix}=0$$which leads to only one independent equality $v_1=3v_2$ among which only $$begin{pmatrix}v_1\v_2end{pmatrix}=begin{pmatrix}{3over sqrt{10}}\{1over sqrt{10}}end{pmatrix}$$has unit norm and the other answers are only a simple factor of this one. Following the same logic for $lambda=30$, we get to $$begin{pmatrix}v_1\v_2end{pmatrix}=begin{pmatrix}{1over sqrt{10}}\-{3over sqrt{10}}end{pmatrix}$$

The zero vector is always a solution to the homogeneous equation. What you’re looking for here are non-zero solutions.

If you performed the row-reduction correctly for the eigenvalue $0$, you should’ve gotten $$begin{bmatrix}1&-3\0&0end{bmatrix}.$$ Using the method described here you can then read from this matrix that $(3,1)^T$ spans its null space, i.e., is an eigenvector of $0$. The situation for the other eigenvalue will be similar: after row-reduction you’ll have a matrix with one zero row from which you can read a basis for its null space. However, there’s a shortcut: this is a real symmetric matrix, so its eigenspaces are mutually orthogonal. This means that an eigenvector of $30$ is $(1,-3)^T$, which is orthogonal to $(3,1)^T$.

In fact, for such a small matrix you can find these eigenvectors and eigenvalues by inspection. The null space of a matrix is the orthogonal complement of its row spaceThe latter is obviously spanned by $(3,-9)^T$, so $(9,3)^T$ is an eigenvector with eigenvalue $0$. Alternatively, you might have spotted that $3times$ the first column plus the second yields the zero vector, which means that $(3,1)^T$ is an eigenvector of zero. You get the other eigenvalues “for free” since their sum is equal to the trace of the matrix, and for an eigenvector, you can either use the symmetry of the matrix as above, or notice that the row space of $30I-A$ is also obviously spanned by its first row.

Once you have an eigenvector for each eigenvalue, simply divide them by the norms to obtain unit vectors. Note, though, that it doesn’t make sense to speak of “the” eigenvectors of a matrix. Any nonzero scalar multiple of an eigenvector is also an eigenvector, so even after normalization, if you have a unit eigenvector $mathbf u$, then $-mathbf u$ is also a unit eigenvector with the same eigenvalue.

A number, $lambda$, is an eigenvalue for linear transformation, A, if and only if there exist a non-zero vector, v, such that $Av= lambda v$. Certainly, yes, the vector v= <0, 0> satisfies $Av= lambda v$. The question is whether or not there exist non-zero vectors. Here, yes, $lambda= 0$ is an eigenvalue. A vector, $begin{pmatrix}x \ y end{pmatrix}$ is an eigenvalue if and only if $begin{pmatrix}3 & -9 \ -9 & 27end{pmatrix}begin{pmatrix}x \ yend{pmatrix}= begin{pmatrix}3x- 9y \ -9x+ 27yend{pmatrix}= begin{pmatrix} 0x \ 0y end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix}$. So we have 3x- 9y= 0, -9x+ 27y= 0. Yes, x= y= 0 is a solution to this but the whole point of an "eigenvalue" is that there are non-zero solutions. From 3x- 9y= 0, we have 3x= 9y and then x= 3y. Replacing x by 3y in the other equation gives 3(3y)- 97= 9y- 9y= 0 so that any vector $begin{pmatrix}3y \ yend{pmatrix}$ is an eigenvector for any y.