# Prove that $(1, i);(1,-i)$ are characteristic vectors of $\begin{bmatrix} a & b \\ -b & a \end{bmatrix}$

Prove that $(1, i);(1,-i)$ are characteristic vectors of $\begin{bmatrix} a & b \\ -b & a \end{bmatrix}$

I've found the polynomial characteristic: $\lambda^2-2a\lambda+a^2+b^2$ and the roots are:

$\lambda_{1} = \frac{a+ib}{\lambda} \\ \lambda_{2} = \frac{a-ib}{\lambda}$

But, I don't know how to resolve the system and find characteristic vectors.

Thanks so much.

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apply the matrix transformation to your vectors. do you get back scaled versions of the input vectors? – akkkk Jun 13 '12 at 22:24

In order to test if a given nonzero vector $\mathbf{v}$ is a characteristic vector (aka eigenvector) of a given matrix $A$, you do not need to find the eigenvalues, or characteristic polynomial of the matrix! All you have to do is compute $A\mathbf{v}$ and see if what you get is a scalar multiple of $\mathbf{v}$ or not.

So you need to compute $$\left(\begin{array}{cc}a&b\\-b&a\end{array}\right)\left(\begin{array}{c}1\\i\end{array}\right) = \left(\begin{array}{c}a+bi\\-b+ai\end{array}\right).$$ Now, is $(a+bi,-b+ai)$ a scalar multiple of $(1,i)$? Yes: check that if $\lambda=a+bi$, then $\lambda(1,i) = (a+bi,-b+ai)$. Since $A(1,i)^t$ is a scalar multiple of $(1,i)$, then $(1,i)$ is a characteristic vector of $A$.

Similarly with $(1,-i)$.

(I'm reminded of how the vast majority of my Calculus students will, when faced with a problem like "Verify that $f(x)=2\sin x-3\cos x$ is a solution to the differential equation $y''+y=0$" will proceed to try to find the general solution to the equation and see if the given $f(x)$ is of that general form, instead of simply plugging in $f$ and checking the equality... even after I tell them to just plug in... then again, if they are asked to verify that $17$ is a solution to $x^2 -27x + 170=0$, they will proceed to solve the quadratic instead of just plugging in, too)

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Thanks so much!.. – Albert Jun 27 '12 at 22:43
Or, "How many real roots does the quadratic $3x^2 - 7x + 15$ have?" Three-fourths of my class would use the full quadratic formula. – Eric Jablow May 16 '13 at 14:40

So you have to check whether $$\begin{pmatrix}a&b\\\!\!\!-b&a\end{pmatrix}\binom{1}{i}=\lambda\binom{1}{i}\,\,,\,\lambda=\lambda_1\,\,or\,\,\lambda_2$$and the same for the other given vector. Well, now just verify that you indeed get the above matrix equation right when you use the values you got for $\,\lambda\,$...and find out which one corresponds to which vector.

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