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Question 2 of 20 View Policies Current Attempt in Progress Find the characteristic equation, the eigenvalues, and bases for the eigenspaces of the following matrix. $A = \begin{bmatrix} 1 & -8 \\ 0 & 1 \end{bmatrix}$ The characteristic equation is $oxed{}$ = 0 The eigenvalue $\lambda$ is $oxed{}$ A basis for the eigenspace is $oxed{\begin{pmatrix} ?\\? \end{pmatrix}}$

          Question 2 of 20
View Policies
Current Attempt in Progress
Find the characteristic equation, the eigenvalues,
and bases for the eigenspaces of the following matrix.
$A = \begin{bmatrix} 1 & -8 \\ 0 & 1 \end{bmatrix}$
The characteristic equation is  $oxed{}$ = 0
The eigenvalue $\lambda$ is $oxed{}$
A basis for the eigenspace is $oxed{\begin{pmatrix} ?\\? \end{pmatrix}}$
        
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Question 2 of 20
View Policies
Current Attempt in Progress
Find the characteristic equation, the eigenvalues,
and bases for the eigenspaces of the following matrix.
A = 
    < b m a t r i x >
The characteristic equation is  oxed = 0
The eigenvalue λ is oxed
A basis for the eigenspace is oxed
    < p m a t r i x >

Added by Candice B.

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Elementary and Intermediate Algebra
Elementary and Intermediate Algebra
Alan S. Tussy, R. David Gustafson 5th Edition
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Question 2 of 20 View Policies Current Attempt in Progress Find the characteristic equation, the eigenvalues, and bases for the eigenspaces of the following matrix. A = 1-8 01 The characteristic equation is The eigenvalue X is A basis for the eigenspace is
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Transcript

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00:01 Hello everyone this question the matrix is given we have to find the eigenvalue and eigenspace for this matrix.
00:11 So for eigenvalue i write this as determinant of minus 1 minus lambda 0 4 minus 1 minus lambda 0 minus 2 0 minus 1 minus lambda equal to 0.
00:26 So evaluating this determinant minus 1 minus lambda into minus 1 minus lambda the whole square another term will become 0.
00:37 So minus 1 minus lambda taking minus common we will be having 1 plus lambda the whole square equal to 0.
00:48 So we will be having minus 1 minus lambda minus into minus will be so 1 plus lambda the whole square equal to 0.
00:55 So this gives you minus 1 minus lambda 1 plus lambda square plus 2 lambda equal to 0.
01:04 So this gives you minus 1 minus lambda square minus 2 lambda minus lambda minus lambda cube minus 2 lambda square equal to 0.
01:15 So this gives you minus lambda cube minus 3 lambda minus 3 lambda square minus 1 equal to 0.
01:23 So this can be rewritten as lambda cube plus 3 lambda plus 3 lambda square plus 1 equal to 0.
01:34 So from this we can found lambda by synthetic division 1 3 3 1.
01:40 So if i take minus 1 i will be having minus 1 2 minus 2 1 minus 1 which is 0.
01:49 So this gives you lambda equal to minus 1 is one root and another root will be lambda square plus 2 lambda plus 1 equal to 0 which can be written as lambda plus lambda plus 1 equal to 0.
02:04 So this gives you lambda plus 1 plus 1 into lambda plus 1.
02:09 So this gives you lambda 1 into lambda 1 equal to 0.
02:13 So this gives the value lambda minus 1 minus 1.
02:16 So we can say that the eigenvalue for this matrix is lambda equal to minus 1.
02:24 So we have to find eigenspace for this lambda.
02:28 Next we have to find the eigenspace for lambda equal to minus 1.
02:36 For this we have to find a minus lambda i that is a is minus 1 0 0 4 minus 1 0 minus 2 0 minus 1...
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