My answer for eigenspace is valid right?

[Rijad Hadzic]

Homework Statement

Determine the characteristic polynomials, eigenvalues, and corresponding eigenspaces of the given 2x2 matricies

Homework Equations

The Attempt at a Solution

$<br /> \begin{pmatrix}<br /> 5 & 2\\<br /> -8 & -3 \\<br /> \end{pmatrix}$
thus

$<br /> \begin{pmatrix}<br /> 5-\lambda & 2\\<br /> -8 & -3-\lambda \\<br /> \end{pmatrix}$

determinant is = to: $\lambda^2 -2\lambda + 1$

which gives value lambda = 1

plugging into $<br /> \begin{pmatrix}<br /> 5-\lambda & 2\\<br /> -8 & -3-\lambda \\<br /> \end{pmatrix}$

you get$<br /> \begin{pmatrix}<br /> 4 & 2\\<br /> -8 & -4 \\<br /> \end{pmatrix}$

using rref you get$<br /> \begin{pmatrix}<br /> 1 & .5\\<br /> 0 & 0\\<br /> \end{pmatrix}$

setting x2 = r, I get eigenspace r*$<br /> \begin{pmatrix}<br /> -1/2\\<br /> 1 \\<br /> \end{pmatrix}$

but my book is telling me the anser is r*$<br /> \begin{pmatrix}<br /> 1\\<br /> -2 \\<br /> \end{pmatrix}$

our answers are the same thing right?

 

[Dick]

Homework Statement

Determine the characteristic polynomials, eigenvalues, and corresponding eigenspaces of the given 2x2 matricies

Homework Equations

The Attempt at a Solution

$<br /> \begin{pmatrix}<br /> 5 & 2\\<br /> -8 & -3 \\<br /> \end{pmatrix}$
thus

$<br /> \begin{pmatrix}<br /> 5-\lambda & 2\\<br /> -8 & -3-\lambda \\<br /> \end{pmatrix}$

determinant is = to: $\lambda^2 -2\lambda + 1$

which gives value lambda = 1

plugging into$<br /> \begin{pmatrix}<br /> 5-\lambda & 2\\<br /> -8 & -3-\lambda \\<br /> \end{pmatrix}$

you get$<br /> \begin{pmatrix}<br /> 4 & 2\\<br /> -8 & -4 \\<br /> \end{pmatrix}$

using rref you get$<br /> \begin{pmatrix}<br /> 1 & .5\\<br /> 0 & 0\\<br /> \end{pmatrix}$

setting x2 = r, I get eigenspacer*$<br /> \begin{pmatrix}<br /> -1/2\\<br /> 1 \\<br /> \end{pmatrix}$

but my book is telling me the anser isr*$<br /> \begin{pmatrix}<br /> 1\\<br /> -2 \\<br /> \end{pmatrix}$

our answers are the same thing right?

Of course they are. If $v$ is an eigenvector then so is $cv$ for any nonzero $c$. Your values of $r$ just differ by a factor of -2.

 

[Rijad Hadzic]

Of course they are. If $v$ is an eigenvector then so is $cv$ for any nonzero $c$. Your values of $r$ just differ by a factor of -2.

alright ty was just making sure. I do overthink small things like this but your explanation makes sense.

 

[Ray Vickson]

Homework Statement

Determine the characteristic polynomials, eigenvalues, and corresponding eigenspaces of the given 2x2 matricies

Homework Equations

The Attempt at a Solution

$<br /> \begin{pmatrix}<br /> 5 & 2\\<br /> -8 & -3 \\<br /> \end{pmatrix}$
thus

$<br /> \begin{pmatrix}<br /> 5-\lambda & 2\\<br /> -8 & -3-\lambda \\<br /> \end{pmatrix}$

determinant is = to: $\lambda^2 -2\lambda + 1$

which gives value lambda = 1

plugging into$<br /> \begin{pmatrix}<br /> 5-\lambda & 2\\<br /> -8 & -3-\lambda \\<br /> \end{pmatrix}$

you get$<br /> \begin{pmatrix}<br /> 4 & 2\\<br /> -8 & -4 \\<br /> \end{pmatrix}$

using rref you get$<br /> \begin{pmatrix}<br /> 1 & .5\\<br /> 0 & 0\\<br /> \end{pmatrix}$

setting x2 = r, I get eigenspacer*$<br /> \begin{pmatrix}<br /> -1/2\\<br /> 1 \\<br /> \end{pmatrix}$

but my book is telling me the anser isr*$<br /> \begin{pmatrix}<br /> 1\\<br /> -2 \\<br /> \end{pmatrix}$

our answers are the same thing right?

You should point out that "1" is a double eigenvalue; that is, the eigenvalues of the matrix are 1,1. You might also want to point out that this matrix is "deficient": its "eigenspace" has only one dimension and so does not span the whole space. (It I were marking this question I would give full points only if the student mentioned those things---but of course, I am not marking it.)

 

[Mark44]

Thread moved to Calculus section. Questions about eigen-<whatever> are well beyond precalculus, IMO.