My answer for eigenspace is valid right?

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The discussion centers on determining the characteristic polynomials, eigenvalues, and corresponding eigenspaces of the matrix \begin{pmatrix} 5 & 2\\ -8 & -3 \end{pmatrix}. The characteristic polynomial is calculated as \lambda^2 - 2\lambda + 1, yielding a double eigenvalue of 1. The eigenspaces derived from the reduced row echelon form (RREF) lead to two equivalent eigenvectors, r*\begin{pmatrix} -1/2\\ 1 \end{pmatrix} and r*\begin{pmatrix} 1\\ -2 \end{pmatrix}, differing by a scalar factor of -2. The matrix is identified as deficient, having a one-dimensional eigenspace.

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Homework Statement


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

Homework Equations

The Attempt at a Solution


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

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

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

which gives value lambda = 1

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

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

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

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

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

our answers are the same thing right?
 
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Rijad Hadzic said:

Homework Statement


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

Homework Equations

The Attempt at a Solution


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

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

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

which gives value lambda = 1

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

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

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

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

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

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.
 
Dick said:
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.
 
Rijad Hadzic said:

Homework Statement


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

Homework Equations

The Attempt at a Solution


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

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

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

which gives value lambda = 1

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

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

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

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

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

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

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