My linear algebra professor gave us a practice worksheet for our upcoming exam, and the answer key he gave has me very confused. I'm not entirely sure his answers are correct, which is why I thought i'd bring the question to you good folks. The question is as follows:

Find a basis for the solution space of Ax = 0 if

$$A = \begin{bmatrix} 1&1&0&0\\ -2&-2&0&0\\ 0&0&1&-1\\ -1&-1&0&1 \end{bmatrix} $$

Reduced row echelon form yields:

$$ \left[ \begin{array}{cccc|c} 1&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&0 \end{array} \right] $$


$x_1 = -x_2$

$x_3 = 0$

$x_4 = 0$

$\implies$ Basis for null(A) = $\begin{bmatrix} -1\\ 1\\0\\0 \end{bmatrix}$

His solution, however, gives:

Basis for null(A) = $\begin{bmatrix} -1\\ 1\\0\\0 \end{bmatrix}$ , $\begin{bmatrix} 0\\ 0\\-1\\1 \end{bmatrix}$

Did I make a mistake here? Or is this answer incorrect. I have always always been under the assumption that $dim(\mathbb R^4) = rank(A) + nullity(A)$ per the rank-nullity theorem. In this case:

$4 = 3 + nullity(A) \implies nullity(A) = 1$ meaning, to me, the null space should contain 1 vector and not 2.

If i'm totally off base here please let me know, as I really want to understand this stuff. Thanks!

  • 2
    $\begingroup$ Did you check your lecturer's solution? Take the vectors $v_1$ and $v_2$ he gave you and see whether $Av_1$ and $Av_2$ are both the zero vector? $\endgroup$ – Angina Seng Jun 27 '17 at 4:43
  • $\begingroup$ It did not give the zero vector for A$v_2$, although I was not aware of that check, thank you for sharing. So i'm guessing this means his solution is wrong? $\endgroup$ – FuegoJohnson Jun 27 '17 at 4:49
  • $\begingroup$ Yes, this means his solution is wrong! $\endgroup$ – Angina Seng Jun 27 '17 at 4:55
  • $\begingroup$ @FuegoJohnson This sanity check is nothing more than applying the definition of a null space to the purported basis of a null space. $\endgroup$ – amd Jun 27 '17 at 6:58
  • $\begingroup$ Quite likely there’s a typo in the matrix $A$. A small change to it will make the null space two-dimensional. $\endgroup$ – amd Jun 27 '17 at 7:03

Wolfram Alpha agrees with you:


(rank 3, meaning only one vector spanning the null space)

  • $\begingroup$ Thanks for confirming. You assume the professor will give the correct answer key, so it's always disorienting when their answer is incorrect. As a a somewhat related question, this is the same logic used for finding the eigenspace of a matrix as well, correct? $\endgroup$ – FuegoJohnson Jun 27 '17 at 4:52
  • $\begingroup$ Can you be more specific please? :) $\endgroup$ – Ofek Gillon Jun 27 '17 at 5:18
  • $\begingroup$ Since the eigenspace is the null space of $det(A - \lambda I)$, it follows that the dimension of the eigenspace is $n - rank(A - \lambda I)$ for any square matrix. Is this correct? $\endgroup$ – FuegoJohnson Jun 27 '17 at 5:56
  • 1
    $\begingroup$ Oh, then of course $\endgroup$ – Ofek Gillon Jun 27 '17 at 5:56

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