# What is the associated eigenvalue?

## Homework Statement

V is an eigenvector of the nxn matrix A, with a eigenvalue of 4. explain why V is a eigenvector of A^2+2A+3I. what is the associated eigenvalue?

## The Attempt at a Solution

is the eigenvalue of A^2+2A+3I=21?

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## Answers and Replies

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quasar987
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What does it mean that V is an eigenvector of A with eigenvalue 4?

Answer that, and then ask yourself what happens when you multiply the matrix (A^2+2A+3I) by V from the right.

The answer for the eigenvalue is not 21.

I don't understand how A^2+2A+3I is populated in what I assume is a 2x2 matrix? If you tell me that it will help me a lot. I don't understand how a polynomial populates a matrix

to answer you question. A is a matrix (nxn) v is a nonzero vector in R^n. Av is a scalar multiple of lamda, Av=landav. lamda is the eigenvalue. the unknown here is A either its a matrix that i do know the vectors of or it's a variable that populates the matrix that is used to calculate the eigenvalue...right? so as you say A^2+2A+3I is the matrix how does it populated matrix...what is its dimensions?

quasar987
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I don't understand your use of the work "populated".

Here, A is an nxn matrix. And as you probably know, multiplying two nxn matrices or adding two nxn matrices, or multiplying an nxn matrix by a scalar gives out another nxn matrix. So (A^2+2A+3I) is an nxn matrix.

Now, what is (A^2+2A+3I)v, knowing that Av=4v?

eigenvalue

looks like it equals 4...this has been helpful. I've had a misconception that coefficients in A^2+2A+3I would be entered into the nxn matrix in some manner..e.g. [[A^2, 2], [3, 0]] or if i knew the vectors of A i could try to calculated the eigenvalue. so is 4 the correct solution?

quasar987
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No, it isn't.

Use the fact that multiplication between matrix and vector is distributive. Meaning that for A, B two nxn matrices and u in R^n, (A+B)u=Au+Bu.

HallsofIvy
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Saying that v is an eigenvector of A with eigenvalue 4 means that Av= 4v.

What is (A2+ 2A+ 3I)v= A(Av)+ 2Av+ v?

eigenvalue

ok i get what you've been saying about using the distributive rule (thanks) A^2v+2Av+3v (is 3v correct 3I*v=3v). so are we at this point A^2v+2Av+3v=4v? if yes then what? thanks for your help.

quasar987
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Why "A^2v+2Av+3v=4v"? What you know is that Av=4v.

So use that to write A^2v+2Av+3v = A(4v)+2(4v)+3v=... (you finish)

HallsofIvy
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ok i get what you've been saying about using the distributive rule (thanks) A^2v+2Av+3v (is 3v correct 3I*v=3v). so are we at this point A^2v+2Av+3v=4v? if yes then what? thanks for your help.
No, I didn't say that at all. Av= 4v. What is A(Av)= A(4v)= 4Av? What is 2Av?

A(4v)+2(4v)+3v=4(Av)+8v+3v=4(Av)+8v+3v=4(4v)+8v+3v=16v+8v+3v=27v does that make sense?

quasar987
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Plenty.

thank you. i'll do better with the next question!