Questions on Calculating Eigenvalues and Eigenvectors

Hi,

in the Calculating Eigenvalues and Eigenvectors, after we solved the characteristic polynomial, then next step is to get the eigenvectors.


I am having issue to understand how to get the x= 1, y = 0, x= 1, y = 1 from these equations.

Also in the practice questions


Since the lambda = 1, 11
Then we can get
4x + 3y=y
9x+4y = x

4x + 3y=11y
9x+4y = 11x

I can only get y = -2x from first set of equations, how do I solve this?

Hi @Bio_J!

To answer the first question, you obtain the answers x=1, y=0 and x=1, y=1 by solving each of the equations and finding the values for x and y.

For the first set of equations:

2x + y = 2x \\ 0x + 3y = 2y

We can simplify the first equation to get

2x + y = 2x => y = 2x - 2x => y = 0

Then, to solve for x, we can plug in y = 0 to the original equations to get:

2x + 0 = 2x => x = x \\ 0x = 0 => 0 = 0

Choosing some non-zero x that satisfies these equations, we can get x = 1. Therefore we have x = 1, y = 0 as the solution for the first set of equations.

For the second set of equations:

2x + y = 3x \\ 0x + 3y = 3y

We can simplify the second equation to get

0x + 3y = 3y => y = y

Then, choosing some non-zero value of y that satisfies this, we can get y = 1.

Next, to solve for x, we can plug in y = 1 to get:

2x + 1 = 3x => 1 = x \\ 0x + 3y = 3y => y = y

So we get that x = 1, y = 1 for the second set of equations.

Note that the solutions for x and y will technically be infinite, as any multiple of the values for x and y will solve the equations. For example, for the second set of equations, (1, 1), (2, 2), (10, 10), etc. would all be solutions. Feel free to plug in some other multiples of the solutions to verify this is the case!

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For your second question. y = -2x is indeed correct for the first set of equations.

We can simplify

4x + 3y = y \\ 9x + 4y = x

to:

4x = -2y => 2x = -y => -2x = y \\ 8x = -4y => 2x = -y => -2x = y

So we get y = -2x.

Any values of x and y that satisfy this equation will be a solution. We can try out some value of x, say x = 1, which would then give us y = -2. Any multiples of (1, -2) would also be a solution, like (-1, 2) which is (1, -2) times -1, which is the solution given in the practice questions.

I hope this helps!

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Thank you for the answer.


Question on the 3*3 matrix eigenvector here.

What does 0=0 means here? If my understanding is correct, the only useful info is x1=2x2-0.5x3, which as long as combination of x1,x2,x3 fits this equaltion, it will work? right?

Hi @Bio_J,

Yes, your understanding seems correct.

The 0 = 0 equations will not give us any information, so the only useful information then is from the second equation -x_1 + 2x_2 - 0.5x_3 = 0.

Because that’s the only equation that needs to be satisfied, then any combination of x1, x2, x3 (where the equation is still true) would work.

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In the next video On the Number of Eigenvectors


The A - identity matix doesn’t seem correct, why in here the numbers changed to its addtive inverse form? should it be the same?

like in example 2, it should be[[2-lambda,0,0],[-1,4-lambda,-0.5],[4,0,2-lambda]]?

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Hi @Bio_J,

Thank you for catching that! You’re right, the signs of those values should not be switched, since the A matrix is not being subtracted.

I’ll put in a request to have that fixed in the future.

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