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$\begin{pmatrix} 2 & 12 \\ 2 & -3 \end{pmatrix} \textbf{v} = k \textbf{v}.$

For what values of $k$ which are real there exists a 2-vector $v$ satisfying the above equation?

How do I start this problem? Should I just let $n=\binom{a}{b}$ and plug it in?

Thanks!

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    Have you heard of eigenvalues and eigenvectors? – 2017-02-12
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    No, are they necessary for this problem? – 2017-02-12
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    This problem is a very basic question on this subject. This is not to say you can't solve it using other, even elementary techniques, but eigenvalue/eigenvectors is the natural way to tackle this problem. – 2017-02-12
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    Ok, but I've never heard of those. Could you guide me through with another form of solution? – 2017-02-12
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    Your own suggested course of action is plausible. Plugging it in will gt you a system of two equation with three variables: that's okay, because if $(k,v)$ is a solution then $(k,t\cdot v)$ is also a solution for all $t \in \mathbb{R}$. – 2017-02-12
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    I don't know how to find k, I have $\binom{a}{b}k=\binom{2a+12b}{2a-3b}$ – 2017-02-12

2 Answers 2

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Let's try doing it via elementary means. Using your notation, we have the system:

\begin{align} \tag{1}\label{eq1}2a+12b &= k \cdot a\\ \tag{2}\label{eq2}2a-3b &= k\cdot b \end{align} With $\eqref{eq1}$, we find that $(k-2)a=12b$ $\,\,(*)$.

Multiplying both sides of $\eqref{eq2}$ by $12$, we get $24a-3\cdot 12b=k\cdot12b$. We now substitute $(*)$ into this equation, obtaining

$$24a-3(k-2)a=k(k-2)a$$

which we may rearrange to obtain $$(k^2+k-30)\cdot a = 0$$

This implies that either $a=0$, $k=5$ or $k=-6$.

We see via $(*)$ that $a=0$ implies $b=0$, and then any $k$ will do. These are the trivial solutions.

Now, suppose $k=5$. Then we get from $(*)$ that

$$3a=12b \iff a=4b$$

So we get a family of solutions $n=(4t,t)$ and $k=5$. For instance, $n=(4,1)$ and $k=5$ is a solution.

Now, suppose $k=-6$. Then we get from $(*)$ that

$$-8a=12b \iff -2a=3b$$

So we get a family of solutions $n=(-3t,2t)$ and $k=-6$. For instance, $n=(-3,2)$ and $k=-6$ is a solution.

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    Thanks for the non eigen solution! Very much appreciated! :) Either way, you need to produce a system of equations with the top and bottom of the vector, correct? – 2017-02-12
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    Yes. Notice that we had to solve a quadratic equation to get the values of $k$. That's because our system has two equations. In geneal, if the system has $n$ equations, we'd have to find the roots of degree $n$ polynomial, and this can become hard fairly quickly. – 2017-02-12
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For a matrix $A$ if $Av=kv$ for $v$ a non zero vector then $k$ is an eigenvalue of $A$ and $v$ an eigenvector. $$ \begin{pmatrix} 2 & 12 \\ 2 & -3 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix}2v_1+12v_2 \\ 2v_1-3v_2 \end{pmatrix} $$ So we want: $$ 2v_1+12v_2=kv_1\\ 2v_1-3v_2=kv_2 $$ This system is equivalent to the following: $$ (2-k)v_1+12v_2=0\\ 2v_1+(-3-k)v_2=0 $$ which corresponds to: $$ \begin{pmatrix} 2-k & 12 \\ 2 & -3-k \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix}0 \\0 \end{pmatrix} $$

This systems has a non trivial solution $v=(v_1,v_2)$ when the determinant of the matrix is $0$, namely when: $$ (2-k)(-3-k)-24=0 \Leftrightarrow k^2+k-30=0 \Leftrightarrow (k+6)(k-5)=0 \Leftrightarrow k=-6, \quad k=5 $$ For each of these values of $k$ you can find $v\neq (0,0)$ that satisfy the above equations.

Can you find such $v$'s for each of these $k$'s? If not, I can elaborate further.

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    Ahh, ok! are the v's $\binom{4}{1}$, $\binom{-3}{2}$ for k=5, -6 respectively? – 2017-02-12
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    I was just learning what an eigenvector was! So you produced a system of equations out of that. – 2017-02-12
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    @user406996 I didn't use anything apart for a system of equations corresponding to the equation you have. I only mentioned for reference the terms eigenvalue and eigenvector. You don't seem to appreciate my answer for what it is! – 2017-02-12
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    Hmm, then I was mislead by your solution, sorry. I'm not quite sure how to use the eigenvalues and vectors, so its a rough ride trying to learn what they are. – 2017-02-12
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    Ohh, now I see how you did it. I saw "system" and I immediately thought you were using system of equations. – 2017-02-12
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    @user406996 It's only doing a bit of algebra with the initial equation given in your post. – 2017-02-13