# Finding eigenvectors for diagonalization

• iamsmooth
In summary: Also, I'm confused why you have 6x-2.5y=0 when it should be x+3y=0, which would mean that y=-1/3x...In summary, to find the eigenvectors for the given matrix A, you need to solve the equations 6x-2.5y=0 and -7x+2.5y=0 for each eigenvalue. This will result in the eigenvectors x[1, 34] and x[-1, 34] for the eigenvalues -2.5 and -6.5 respectively. From there, you can construct the invertible matrix S and diagonal matrix D to satisfy the given equation.
iamsmooth

## Homework Statement

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

Find an invertible S and a diagonal D such that $$S^{-1}AS=D$$

...

## The Attempt at a Solution

So first I need to get eigenvalues so I can get the eigenvectors which will give me the invertible S (or so I believe).

Anyways, here is my work so far:

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

$$(\lambda+6)(\lambda+3)-1.75$$

$$\lambda^2+3\lamda+6\lambda+18-1.75$$

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

Using quadratic equation or factoring, the roots are:

$$\lambda=-2.5$$

$$\lambda=-6.5$$

From here, subbing lambda in, I get:

$$\left[ \begin{array}{cc} -2.5+6 & -0.25 \\ -7 & -2.5+3 \end{array} \right] \left[ \begin{array}{cc} X_1 \\ X_2 \\ \end{array} \right]$$

$$\left[ \begin{array}{cc} -6.5+6 & -0.25 \\ -7 & -6.5+3 \end{array} \right] \left[ \begin{array}{cc} X_1 \\ X_2 \end{array} \right]$$

Solving for these systems I get:

$$E_{-2.5} = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$$

$$E_{-6.5} = \left[ \begin{array}{cc} 1 & 0.5 \\ 0 & 0 \end{array} \right]$$

From here, how do I get the eigenvectors? I think I'm close (if my work is correct so far)

Last edited:
I guess my eigenspace is probably wrong, can someone help me figure that out? If I get that, I know how to get the diagonal.

Thanks.

So I used a matrix calculator to get the vectors (since I can't get them) and it gave me 1, -2 for $\lambda$ of -6.25, and -2, 14 for $\lambda$ of-2.5

So with this, the eigenspace should be

$$S = \left[ \begin{array}{cc} 1 & 1 \\ -2 & 14 \end{array} \right]$$

And when I calculate it out as $S^{-1}AS = D$

$$D = \left[ \begin{array}{cc} -104 & 0 \\ 0 & -40 \end{array} \right]$$

Which seems like it works since it's now diagonal, but the answer is wrong for some reason. I've tried switching the signs and it's still wrong. Really not sure what I'm doing wrong. This is the last question I have for practice and this is getting frustrating. I'm starting to think maybe there's an error with the program.

iamsmooth said:

## Homework Statement

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

Find an invertible S and a diagonal D such that $$S^{-1}AS=D$$

...

## The Attempt at a Solution

So first I need to get eigenvalues so I can get the eigenvectors which will give me the invertible S (or so I believe).

Anyways, here is my work so far:

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

$$(\lambda+6)(\lambda+3)-1.75$$

$$\lambda^2+3\lamda+6\lambda+18-1.75$$

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

Using quadratic equation or factoring, the roots are:

$$\lambda=-2.5$$

$$\lambda=-6.5$$

From here, subbing lambda in, I get:

$$\left[ \begin{array}{cc} -2.5+6 & -0.25 \\ -7 & -2.5+3 \end{array} \right] \left[ \begin{array}{cc} X_1 \\ X_2 \\ \end{array} \right]$$

$$\left[ \begin{array}{cc} -6.5+6 & -0.25 \\ -7 & -6.5+3 \end{array} \right] \left[ \begin{array}{cc} X_1 \\ X_2 \end{array} \right]$$

Solving for these systems I get:

$$E_{-2.5} = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$$

$$E_{-6.5} = \left[ \begin{array}{cc} 1 & 0.5 \\ 0 & 0 \end{array} \right]$$

From here, how do I get the eigenvectors? I think I'm close (if my work is correct so far)
Your error is in subtracting the two different eigenvalues from the same matrix. To find an eigenvalue corresponding to x= -2.5 Look at the equation
$$\begin{bmatrix}-6 & .25 \\ 7 & -3\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= -2.5\begin{bmatrix}x \\ y\end{bmatrix}$$
or, equivalently
$$\begin{bmatrix}6+ 2.5 & -.25 \\ -7 & 3+ 2.5\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}0 \\ 0\end{bmatrix}$$
That reduces to the two equations 6x- 2.5y= 0 and -7x+ 2.5y= 0 which both reduce to y= 34x. Any eigenvector, corresponding to eigenvalue -2.5, must be of the form
$$x\begin{bmatrix}1 \\ 34\end{bmatrix}$$
and you can take x to be 1, say, to get a specific eigenvalue, or $1/\sqrt{34^2+ 1}$ for a unit eigenvalue.

Similarly, to find an eigenvector corresponding to eigenvalue -6.5, solve
$$\begin{bmatrix}-6 & .25 \\ 7 & -3\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= -6.5\begin{bmatrix}x \\ y \end{bmatrix}$$
or, equivalently,
$$\begin{bmatrix}-6+6.5 & .25 \\ 7 & -3+ 6.5\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix} 0.5 & .25\\ 7 & 3.5 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$$

HallsofIvy said:
$$\begin{bmatrix}6+ 2.5 & -.25 \\ -7 & 3+ 2.5\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}0 \\ 0\end{bmatrix}$$
That reduces to the two equations 6x- 2.5y= 0 and -7x+ 2.5y= 0 which both reduce to y= 34x. Any eigenvector, corresponding to eigenvalue -2.5, must be of the form
$$x\begin{bmatrix}1 \\ 34\end{bmatrix}$$
and you can take x to be 1, say, to get a specific eigenvalue, or $1/\sqrt{34^2+ 1}$ for a unit eigenvalue.

I don't understand where you get the two equations from that matrix. Even if I accept it, I don't understand how it reduces to y=34x.

## 1. What are eigenvectors and why are they important in diagonalization?

Eigenvectors are special vectors that do not change direction when multiplied by a given matrix. In diagonalization, eigenvectors are used to transform a matrix into a diagonal form, which simplifies calculations and makes it easier to solve problems related to the matrix.

## 2. How do you find eigenvectors for diagonalization?

To find eigenvectors for diagonalization, you first need to find the eigenvalues of the matrix. Then, for each eigenvalue, you solve the equation (A-λI)x = 0, where A is the matrix, λ is the eigenvalue, and x is the eigenvector. This will give you a set of eigenvectors that can be used for diagonalization.

## 3. Can a matrix have more than one eigenvector?

Yes, a matrix can have multiple eigenvectors. In fact, in most cases, a matrix will have more than one eigenvector. Each eigenvector corresponds to a different eigenvalue, and together they form a basis for the vector space.

## 4. What is the significance of the diagonal elements in a diagonalized matrix?

The diagonal elements in a diagonalized matrix represent the eigenvalues of the original matrix. This is significant because the eigenvalues determine the behavior of the matrix, such as whether it is invertible or not, and how it will affect other matrices in calculations.

## 5. Are eigenvectors and eigenvalues always real numbers?

No, eigenvectors and eigenvalues can be complex numbers. This usually happens when the matrix has complex elements. However, if all the elements in the matrix are real numbers, then the eigenvectors and eigenvalues will also be real numbers.

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