# eigenvalues/eigenvectors of a 3x3 matrix

#### eigenvectorisnotfun

i have a 3x3 matrix

$$\displaystyle \begin{pmatrix}-2 & -8 & -12\\1 & 4 & 4\\0 & 0 & 1\end{pmatrix}$$

i got the eigenvalues of 2, 1, and 0.
im having a big problem with how to get the corresponding eigenvectors

if anyone can help me that would be great!

#### AllanCuz

i have a 3x3 matrix

$$\displaystyle \begin{pmatrix}-2 & -8 & -12\\1 & 4 & 4\\0 & 0 & 1\end{pmatrix}$$

i got the eigenvalues of 2, 1, and 0.
im having a big problem with how to get the corresponding eigenvectors

if anyone can help me that would be great!
For every eigenvalue there is a corresponding eigenvector. So we need to find our eigenvectors 3 times!

How we do this is place lambda on the diagonal in the matrix! In other words,

$$\displaystyle \begin{pmatrix} {-2 - \lambda }& -8 & -12\\ 1 & { 4- \lambda } & 4\\0 & 0 & {1 - \lambda } \end{pmatrix}$$

Let's pick $$\displaystyle \lambda = 0$$

$$\displaystyle \begin{pmatrix} {-2}& -8 & -12\\ 1 & { 4 } & 4\\0 & 0 & {1} \end{pmatrix}$$

Now we need to row reduce!

$$\displaystyle 12 R_3 + R_1 --> R_1$$

$$\displaystyle \begin{pmatrix} {-2}& -8 & 0\\ 1 & { 4 } & 4\\0 & 0 & {1} \end{pmatrix}$$

$$\displaystyle R_2 - 4R_3 --> R_2$$

$$\displaystyle \begin{pmatrix} {-2}& -8 & 0\\ 1 & { 4 } & 0\\0 & 0 & {1} \end{pmatrix}$$

$$\displaystyle 2 R_2 + R_1 --> R_1$$

$$\displaystyle \begin{pmatrix} {0}& 0 & 0\\ 1 & { 4 } & 0\\0 & 0 & {1} \end{pmatrix}$$

At this point I don't think we can reduce any more.

Now we know there exists a set of vectors such that

$$\displaystyle \begin{pmatrix} {0}& 0 & 0\\ 1 & { 4 } & 0\\0 & 0 & {1} \end{pmatrix} \begin{pmatrix} V_1 \\ V_2 \\V_3 \end{pmatrix} = 0$$

Carry out the multiplication and find your values. There might be trivial solutions but that's okay!

Find more at: Eigenvalues and Eigenvectors

• eigenvectorisnotfun

#### HallsofIvy

MHF Helper
Or, use the definition of "eigenvalue" and "eigenvector".

If you are right that 2 is an eigenvalue, then there must exist a vector, $$\displaystyle \begin{pmatrix}x \\ y \\ z\end{pmatrix}$$, x, y, z not all 0, such that
$$\displaystyle \begin{pmatrix}-2 & -8 & -12 \\ 1 & 4 & 4 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}$$$$\displaystyle = \begin{pmatrix}-2x- 8y- 12z \\ x+ 4y+ 4z \\ z\end{pmatrix}$$$$\displaystyle = \begin{pmatrix}2x \\ 2y \\ 2z\end{pmatrix}$$.

That gives the three equations -2x-8 y- 12z= 2x, x+ 4y+ 4z= 2y, and z= 2z.
Of course, that last equations tells us that z= 0 so the first two become -2x- 8y= 2x, or 8y= -4x so that x= -2y, and x+ 4y= 2y and, again, x= -2y. The fact that there exist an infinite number of solutions to those equations confirms that 2 is an eigenvalue. Any eigenvector corresponding to eigenvalue 2 is of the form $$\displaystyle \begin{pmatrix}-2y \\ y \\ 0\end{pmatrix}= y\begin{pmatrix}-2 \\ 1 \\ 0\end{pmatrix}$$.

Now, to find the eigenvectors corresponding eigenvalues 1 and 0, just solve -2x- 8y- 12z= x, x+ 4y= y, z= z, and -2x- 8y- 12z= 0, x+ 4y= 0, z= 0.

On the contrary, eigenvectors are great fun! (Not to mention extremely useful.)

• eigenvectorisnotfun

#### AllanCuz

Or, use the definition of "eigenvalue" and "eigenvector".

If you are right that 2 is an eigenvalue, then there must exist a vector, $$\displaystyle \begin{pmatrix}x \\ y \\ z\end{pmatrix}$$, x, y, z not all 0, such that
$$\displaystyle \begin{pmatrix}-2 & -8 & -12 \\ 1 & 4 & 4 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}$$$$\displaystyle = \begin{pmatrix}-2x- 8y- 12z \\ x+ 4y+ 4z \\ z\end{pmatrix}$$$$\displaystyle = \begin{pmatrix}2x \\ 2y \\ 2z\end{pmatrix}$$.

That gives the three equations -2x-8 y- 12z= 2x, x+ 4y+ 4z= 2y, and z= 2z.
Of course, that last equations tells us that z= 0 so the first two become -2x- 8y= 2x, or 8y= -4x so that x= -2y, and x+ 4y= 2y and, again, x= -2y. The fact that there exist an infinite number of solutions to those equations confirms that 2 is an eigenvalue. Any eigenvector corresponding to eigenvalue 2 is of the form $$\displaystyle \begin{pmatrix}-2y \\ y \\ 0\end{pmatrix}= y\begin{pmatrix}-2 \\ 1 \\ 0\end{pmatrix}$$.

Now, to find the eigenvectors corresponding eigenvalues 1 and 0, just solve -2x- 8y- 12z= x, x+ 4y= y, z= z, and -2x- 8y- 12z= 0, x+ 4y= 0, z= 0.

On the contrary, eigenvectors are great fun! (Not to mention extremely useful.)
Is this typically a faster approach then the row reductions I have above?

#### eigenvectorisnotfun

now i finally got it!
what i did wrong was I did not really get that (A-I lambda) = 0
thanks to you both, you both were really of great help (Clapping)