I am functioning on a homework assigntment for my linear algebra class and also i am stumped top top this pesky question which is together follows:

Show the λ is one eigenvalue the A and find one eigenvector, x, equivalent to this eigenvalue.$$A=\beginbmatrix6 & 6\\6 & -3\endbmatrix,\qquad \lambda=-6.$$

In my attempts I:

a) worn down to find $A-6I$ (I being the identity matrix for $2\times2$ matrix)

b) The result of the over gave me the procession :$$\beginbmatrix12 & 6\\6 & 3\endbmatrix$$

From i beg your pardon i said that due to the fact that Column 2 is 2x obelisk 1 it is linear independent which implies null space is no zero. Now i am lost and also do not know what to do. More so im not certain what to carry out next. Mine textbook go an example comparable to this however i do not recognize what measures it takes after this. Any suggestions , hints and helpful input is significantly appreciated :)

Thankyou

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

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edited Mar 27 "14 at 1:50

Mhenni Benghorbal
request Mar 27 "14 in ~ 1:34

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

$$\lambda ^2-3 \lambda -54 = 0 \implies (\lambda +6)(\lambda -9) = 0 \implies\lambda_1 = -6, ~ \lambda_2 = 9$$

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

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

The rref the this is:

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

This provides us an eigenvector of:

$$v_1 = (-1, 2)$$

Of course, there space other possible choices for the eigenvector.

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edited Mar 27 "14 at 2:04
answer Mar 27 "14 at 1:38

AmzotiAmzoti
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$\begingroup$

$$\beginbmatrix6 & 6 \\6 & -3\endbmatrix$$

The definition of the eigenvalue problem is together follows:

$$Ax = \lambda x \rightarrow \left(A - \lambda \inter-base.netbbI\right)x = 0$$

For the to have actually non-trivial solutions, the determinant of $A - \lambda \inter-base.netbbI$ have to be zero--which basically just means subtracting $\lambda$ from each of the diagonals and setup the determinant come zero:

$$det\left(\beginbmatrix6 - \lambda & 6 \\6 & -3 - \lambda\endbmatrix\right) = (6 - \lambda)(-3 - \lambda) - 36 = 0 \\-18 - 6\lambda + 3\lambda + \lambda^2 - 36 = 0 \\\lambda^2 - 3\lambda - 54 = (\lambda - 9)(\lambda + 6) = 0 \\\lambda = 9, -6$$

Now girlfriend go ago to discover the eigenvectors by addressing each difficulty individually.

$\lambda = -6$:

$$\beginbmatrix12 & 6 \\6 & 3\endbmatrix\beginbmatrixx \\y\endbmatrix = 0 \\12x + 6y = 0 \rightarrow y = -2x$$

All this way is the the eignenvector is any kind of vector such that the "y" component is the an adverse of twice the x, such as:

$$\beginbmatrix1 \\-2\endbmatrix$$

It"s specifically easy to discover the eigenvector because it"s a 2$\times$2 matrix. Had actually it to be a bigger matrix, girlfriend would need to actually go v the measures of obtaining the matrix into reduced-echelon kind (except because that one row--usually). The just thing that could make these troubles more complicated is once you have actually a multiplicity (or degeneracy) in the eigenvalues which means there room multiple unique eigenvectors with the exact same eigenvalue.