# How to find eigenvectors of a matrix

• iamsmooth
In summary: If we take x= 2, then y= 28 and <2, 28> is also an eigenvector corresponding to eigenvalue -2.5. Similarly, we can find eigenvectors corresponding to the eigenvalue -6.5. The only problem is that you can't find the values x= 1, 2, whatever, without knowing what y is. But you can choose, say, x= 1, then y= 14 so <1, 14> is a vector. Now, choose x= 2, y= 28 so <2, 28> is another eigenvector. Those eigenvectors are not unique. The "
iamsmooth
My textbook doesn't seem to explain it clearly enough for me to comprehend. But from what I can see, after getting the eigenvalues, you sub them back into the lambdas that are in the matrix:

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

From here, you can solve for the system of equations with Gaussian elimination. After you solved it, the eigenvector is determined by the values of each unknown, where they take the coefficient of the parameters (if present).

(quick example) So if I get a matrix and reduce it to RRE that says:

[ 1 0 2 ]
[ 0 1 3 ]
[ 0 0 0 ]

The the values are:

X3 = t
X2 = 3t
X1 = 2t

So the vectorspace I get from this is:
[ 2 ]
[ 3 ]
[ 1 ]

Or something?

This is probably way wrong. Can someone please help explain to me how to calculate eigenvectors? I need it to solve one of my problems.** I will personally send Christmas cards to anyone who helps me with understanding this (provided that you pm me your address)**

Last edited:
Better yet, I'll post the example:

$$A = \left[ \begin{array}{cc} -6 & 0.25 \\ 7 & -3 \end{array} \right] = \left[ \begin{array}{cc} \lambda+6 & -0.25 \\ -7 & \lambda+3 \end{array} \right]$$

$$\lambda^2+9\lambda+16.25$$

$$\lambda=-2.5$$

$$\lambda=-6.5$$

From here is where I need $(\lambda I - A)x = 0$, which yields:

$$\left[ \begin{array}{cc} -2.5+6 & -0.25 \\ -7 & -2.5+3 \end{array} \right]$$

Now, I have tried using Gaussian elimination, but I get really screwy decimal numbers which I really don't think is right. Can someone take me through the process of calculating it and explain the steps and reasoning? I've checked online eigenvector calculators and it said the eigen vectors are:

(1, -2) for -6.25
(1, 14) for -2.25

Not sure how to get to those results (or if they're even correct).

Thanks.

I have always tended to go back to the definition. As iamsmooth says, $$A= \begin{bmatrix}-6 & .25 \\ 7 & -3\end{bmatrix}$$ has eigenvalues -2.5 and -6.5.

Now, the definition of "eigenvalue" is that it is a number, $\lambda$ such that $Av= \lambda v$ is satisfied by some non-zero vector v. (And, since v= 0 is obviously a solution, it there exists non-zero solutions, there must exist an infinite number of them- the set of all eigenvectors corresponding to a given eigenvector forms a subspace.)

That is, any eigenvector corresponding to eigenvalue -2.5 must satisfy
$$\begin{bmatrix}-6 & .25 \\ 7 & -3\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= -2.5\begin{bmatrix}x \\ y\end{bmatrix}$$

[tex]\begin{bmatrix}-6x+ .25y \\ 7x- 3y\end{bmatrix}= \begin{bmatrix}-2.5x \\ -2.5y\end{bmatrix}[/itex]
That gives the two equations -6x+ .25y= -2.5x and 7x- 3y= -2.5y. Adding 6x to both sides of first equation gives .25y= 3.5x or y= 14x. Adding 3y to both sides of the second equation gives 7x= .5y or, again, y= 14x.

Those two equations are dependent and we cannot solve for specific values of x and y because, as I said, there exist an infinite number of eigenvectors for each eigenvalue. We can choose x to be whatever we want and solve for y, getting an eigenvector. For example, if we take x= 1, then y= 14 so <1, 14>, as iamsmooth got, is an eigenvector corresponding to eigenvalue -2.5.

Last edited by a moderator:

## 1. How do I know if a matrix has eigenvectors?

To determine if a matrix has eigenvectors, we can use the characteristic equation det(A-λI)=0, where A is the given matrix, λ is an eigenvalue, and I is the identity matrix. If the characteristic equation has at least one solution, then the matrix has at least one eigenvector.

## 2. Can I find eigenvectors for any type of matrix?

Eigenvectors can be found for square matrices, which means the number of rows is equal to the number of columns. Rectangular matrices do not have eigenvectors.

## 3. How do I find the eigenvectors of a 2x2 matrix?

To find the eigenvectors of a 2x2 matrix, we first need to find the eigenvalues by solving the characteristic equation det(A-λI)=0. Once we have the eigenvalues, we can substitute them back into the equation (A-λI)v=0, where v is the eigenvector. This will give us a system of equations that we can solve to find the eigenvector(s).

## 4. Is there a formula for finding eigenvectors?

Yes, there is a formula for finding eigenvectors. Once we have the eigenvalues, we can use the formula (A-λI)v=0, where A is the given matrix, λ is an eigenvalue, and v is the eigenvector. This formula will give us a system of equations that we can solve to find the eigenvector(s).

## 5. Can I have more than one eigenvector for a single eigenvalue?

Yes, it is possible to have more than one eigenvector for a single eigenvalue. This is because eigenvectors are not unique and can be scaled by any non-zero constant and still be valid eigenvectors. So, if we find one eigenvector for an eigenvalue, we can multiply it by any non-zero constant to get another eigenvector for the same eigenvalue.

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