I am functioning on a homework-related assigntment for my straight algebra course and i am stumped on this pesky question which is as follows:

Sjust how that λ is an eigenvalue of A and also discover one eigenvector, x, corresponding to this eigenvalue.$$A=eginbmatrix6 & 6\6 & -3endbmatrix,qquad lambda=-6.$$

In my attempts I:

a) worn down to find $A-6I$ (I being the identification matrix for $2 imes2$ matrix)

b) The result of the over offered me the matrix :$$eginbmatrix12 & 6\6 & 3endbmatrix$$

From which i said that because Pillar 2 is 2x column 1 it is straight independent which suggests null space is non zero. Now i am shed and also do not recognize what to carry out. More so im not certain what to execute next. My textbook does an instance similar to this yet i perform not understand what measures it takes after this. Any suggestions , clues and also beneficial input is substantially appreciated :)

Thankyou

You are watching: Show that λ is an eigenvalue of a and find one eigenvector v corresponding to this eigenvalue.

linear-algebra

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edited Mar 27 "14 at 1:50

Mhenni Benghorbal

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asked Mar 27 "14 at 1:34

KingShahmooKingShahmoo

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$endgroup$

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$egingroup$

The characteristic polynomial is provided by $|A - lambda I| = 0$, hence:

$$lambda ^2-3 lambda -54 = 0 means (lambda +6)(lambda -9) = 0 implieslambda_1 = -6, ~ lambda_2 = 9$$

The eigenvectors are discovered by $v_i = 0$. For $lambda_1 = -6$, we have

$$eginbmatrix 12 & 6\ 6 & 3\ endbmatrixv_1 = 0$$

$$eginbmatrix 1&dfrac12\0&0\ endbmatrixv_1 = 0$$

This offers us an eigenvector of:

Of course, tright here are various other feasible options for the eigenvector.

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edited Mar 27 "14 at 2:04

answered Mar 27 "14 at 1:38

AmzotiAmzoti

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$egingroup$

Is this your matrix?

$$eginbmatrix6 & 6 \6 & -3endbmatrix$$

The meaning of the eigenvalue difficulty is as follows:

$$Ax = lambda x ightarrow left(A - lambda inter-base.netbbI ight)x = 0$$

For that to have non-trivial services, the determinant of $A - lambda inter-base.netbbI$ need to be zero--which basically simply suggests subtracting $lambda$ from each of the diagonals and establishing the determinant to zero:

$$detleft(eginbmatrix6 - lambda & 6 \6 & -3 - lambdaendbmatrix ight) = (6 - lambda)(-3 - lambda) - 36 = 0 \-18 - 6lambda + 3lambda + lambda^2 - 36 = 0 \lambda^2 - 3lambda - 54 = (lambda - 9)(lambda + 6) = 0 \lambda = 9, -6$$

Now you go back to uncover the eigenvectors by resolving each trouble individually.

$lambda = -6$:

$$eginbmatrix12 & 6 \6 & 3endbmatrixeginbmatrixx \yendbmatrix = 0 \12x + 6y = 0 ightarrow y = -2x$$

All this indicates is that the eignenvector is any type of vector such that the "y" component is the negative of twice the x, such as:

$$eginbmatrix1 \-2endbmatrix$$

It"s specifically easy to uncover the eigenvector because it"s a 2$ imes$2 matrix. Had it been a larger matrix, you would need to actually go with the procedures of obtaining the matrix right into reduced-echelon create (except for one row--usually). The just point that could make these problems more facility is when you have actually a multiplicity (or degeneracy) in the eigenworths which means tbelow are multiple unique eigenvectors via the same eigenworth.