I just want to check my logic. Am I missing something important?

Let $N$ and $M$ be $d\times d$ matrices, and $\mathbf{u}_1, \mathbf{u}_2, \ldots \mathbf{u}_d$ be a set of linearly independent vectors. I want to show that if $N\mathbf{u}_i = M\mathbf{u}_i$ for all $i\in \{1, 2, \ldots, d \}$, then $N = M$.

Since $N\mathbf{u}_i = M\mathbf{u}_i, \implies N\mathbf{u}_i - M\mathbf{u}_i = \mathbf{0} \implies (N-M)\mathbf{u}_{i} = \mathbf{0} $.

Since $(N-M)\mathbf{u}_i = \mathbf{0}$, each $\mathbf{u}_i$ is in the nullspace of $(N-M)$. But all $\mathbf{u}_i, \mathbf{u}_j$ are linearly independent, meaning the collection $\mathbf{u}_1, \mathbf{u}_2, \ldots \mathbf{u}_d$ spans the nullspace of $(N-M)$, meaning $Nullity(N-M) = d$. By rank-nullity then, $rank(N-M) = 0$ whence $N-M = 0 \implies N=M$.

Please let me know if I've missed an important technicality.

  • $\begingroup$ @Itay4 The question says $d \times d$ matrices, so I think it's safe to assume the background space is $\mathbb{R^{d}}$; RMurphy can you confirm? $\endgroup$ Mar 31, 2017 at 17:15
  • $\begingroup$ Does the proof work over general spaces? But yes, I would be satisfied to show this result on d dimmensional complex or real numbers. $\endgroup$
    – RMurphy
    Mar 31, 2017 at 17:36
  • $\begingroup$ Well, if your space has dimension $d+1$, say, what is a $d \times d$ matrix? $\endgroup$ Mar 31, 2017 at 17:48
  • $\begingroup$ Oh, that's what you meant. I thought you were curious about the nature of the elements (reals, complex, etc), not the dimension. Yes, I was trying to say this result should hold over any vector space $V$ of dimension $d$. That is, $N$ and $M$ are linear maps from $V$ into $V$. $\endgroup$
    – RMurphy
    Mar 31, 2017 at 18:21
  • $\begingroup$ Ah, good! Then, yes, the theorem holds in this case (the proof is the same). $\endgroup$ Mar 31, 2017 at 22:46

1 Answer 1


The core of the proof is correct. There's one tiny point I want to check you understand: when you said the nullity of $N-M$ is exactly $d$, to be clear: in general if a vector space contains $d$ linearly independent vectors, it only follows that its dimension is at least $d$. Of course, in this case it is impossible for the dimension of the null space to be larger than $d$, because the dimension of the space it lives inside is $d$! Does that make sense?

In fact, this line of thinking shows that you don't really need rank nullity: any set of $d$ linearly independent vectors in $\mathbb{R^{d}}$ must also span $\mathbb{R^{d}}$. So let $v$ be an arbitrary vector in $\mathbb{R^{d}}$ - express it as a linear combination of the $u_{i}$, apply $N-M$ et voila, $N-M=0$ identically.

  • $\begingroup$ Thanks for your discussion about the dimension being at least $d$. And thanks for your simpler proof. I understand both. $\endgroup$
    – RMurphy
    Mar 31, 2017 at 17:42

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