Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is from Hoffman & Kunze:

Consider $Ax = 0$ with $A = \bigl(\begin{smallmatrix} a&b\\ c&d \end{smallmatrix} \bigr)$. Such that $ad - bc = 0$ With some element of A nonzero.

Then there is a solution $(x^0_1, x^0_2) $ such that $(x_1,x_2)$ is a solution if and only if there is a scalar $y$ such that $x_1 = yx_1^0$ and $x_2 = yx_2^0$.

Some stuff I've noticed

  1. $ad-bc=0$ is the determinant of the matrix.
  2. if $ad = 0$ then $bc = 0$ so there is at least the cases $ad=0$ & $bc = 0$ or $ad\neq0$ and $bc = ad$

Unfortunately however I am not seeing how to setup the problem for a proof.

Can anybody out there provide a hint as to how to proceed?

--Thanks In Advance.

EDIT: Is it enough to say without loss of generality assume $a \neq 0$ and then $\bigl(\begin{smallmatrix} a&b\\ c&d \end{smallmatrix} \bigr)$ can be transformed to $\bigl(\begin{smallmatrix} 1&\frac{b}{a}\\ c&d \end{smallmatrix} \bigr)$ can be transformed to $\bigl(\begin{smallmatrix} 1&\frac{b}{a}\\ c-c&d-\frac{bc}{a} \end{smallmatrix} \bigr)$ which equals $\bigl(\begin{smallmatrix} 1&\frac{b}{a}\\ 0&0 \end{smallmatrix} \bigr)$ by the assumptions made. so $x_1 = -\frac{b}{a}x_2$ and multiplication by $y$ does not change the fact that it is a solution? Is this enough to solve the whole problem or is there other cases I need to check?

share|improve this question
    
@amWhy fixed thanks. –  AvatarOfChronos Feb 1 '13 at 2:34
add comment

1 Answer 1

up vote 2 down vote accepted

If one of the entries is non-zero, then $A$ cannot be the zero transformation. Therefore the dimension of the range is at least $1$. If the determinant is $0$, then the dimension of the range is less than $2$. Therefore it is equal to $1$. Together this says that there are two linearly independent vectors, $v_1$, $v_2$, such that $Av_1$ is non-zero and $Av_2$ is the zero vector. Since these two vectors are linearly independent, any vector in the domain is a linear combination of them. Thus $Av=0$ if an only if $v=kv_2$ for some $k$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.