I will display Δ

I will display Σ

I will display Δ

https://www.khanacademy.org/math/linear-algebra/alternate-bases/eigen-everything/v/linear-algebra-example-solving-for-the-eigenvalues-of-a-2x2-matrix

At the end of the video
Sal gets the factors (?-5)(?+1),
but says that the eigenvalues are 5 and -1.
Why did the signs change?

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Example 1 Find the eigenvalues and eigenvectors of the following matrix. A = 2,7, - 1, -6

that is:
a 2 x 2 matrix where the rows read:
2 7
and
-1 -6

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Solution

1. The first thing that we need to do is find the eigenvalues. That means we need the following matrix,

In particular we need to determine where the determinant of this matrix is zero.

to determine where the determinant of this matrix is zero.

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So, it looks like we will have two simple eigenvalues for this matrix, and . We will now need to find the eigenvectors for each of these.

Also note that according to the fact above, the two eigenvectors should be linearly independent.

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To find the eigenvectors we simply plug in each eigenvalue into (2) and solve. So, let’s do that.

simply plug in each eigenvalue into (2) and solve.

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In this case we need to solve the following system.

Recall that officially to solve this system we use the following augmented matrix.

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Upon reducing down we see that we get a single equation that will yield an infinite number of solutions.

a single equation that will yield an infinite number of solutions

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Recall that we picked the eigenvalues so that the matrix would be singular and so we would get infinitely many solutions.

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Notice as well that we could have identified this from the original system.

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This won’t always be the case, but in the 2 x 2 case we can see from the system that one row will be a multiple of the other and so we will get infinite solutions.

From this point on we won’t be actually solving systems in these cases.

We will just go straight to the equation and we can use either of the two rows for this equation.

Now, let’s get back to the eigenvector, since that is what we were after.

In general then the eigenvector will be any vector that satisfies the following,

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To get this we used the solution to the equation that we found above.

We really don’t want a general eigenvector however so we will pick a value for to get a specific eigenvector.

We can choose anything (except ), so pick something that will make the eigenvector “nice”.

Note as well that since we’ve already assumed that the eigenvector is not zero THEN - we must choose a value that will not give us zero,

which is why we want to avoid in this case. Here’s the eigenvector for this eigenvalue.

Now we get to do this all over again for the second eigenvalue.

We’ll do much less work with this part than we did with the previous part. We will need to solve the following system.

Clearly both rows are multiples of each other and so we will get infinitely many solutions.

We’ll run with the first because to avoid having too many minus signs floating around.

Doing this gives us,

Note that we can solve this for either of the two variables.

However, with an eye towards working with these later on let’s try to avoid as many fractions as possible.

The eigenvector is then,

Summarizing we have,

Note that the two eigenvectors are linearly independent as predicted.

Example 2 Find the eigenvalues and eigenvectors of the following matrix.