Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This came up in a problem I was working on.

Problem:Let $V$ be an $n$ dimensional vector space over a field $F$. Let $T:V\rightarrow V$ be a linear operator and let $W$ be a $T$ invariant subspace of $V$. Prove or give a counter example to the following:

Let $U$ be another subspace of $V$ such that $V=W\oplus U$. Then $U$ is also $T$ invariant.

Solution Well this is false. I came up with a counter example after spending an embarrassing amount of time guessing and checking. Consider the operator $$ \left(\begin{array}{ccc}1 & -1 & \\0 & 1 &\end{array}\right)$$

This does the job because $V=\langle e_1 \rangle \oplus \langle e_2 \rangle$ and the last summand isn't $T$ invertible.

My question is I feel like I've spent too much time on this. I absolutely used no linear algebra knowledge to come up with this. I remotely remember my professor working on $\mathbb{R}^2$ to pull a counter example out of his hat on an unrelated problem. So I just kept plugging away numbers until something worked. Even my grandmother who evidently has never taken linear algebra could have done that. So what I ask is: How would you smart people do this problem?. How do you come up with a counter example for this particular problem. Can you explain the Linear Algebra behind your thought process?. What geometric intuition helps here? (and in general for counterexamples of this nature).

I hope my question makes sense.

Thank you for your comments and answers.

share|cite|improve this question
The statement is generally false because the choice of $U$ isn't unique. You should note that the freedom of choosing $U$ is quite arbitrary. For example, consider the following matrix $\left[\begin{array}[rr]\ 1&0\\0&2\end{array}\right]$. $\langle e_1\rangle$ is an invariant subspace, so is $\langle e_2\rangle$. Well, how about choosing another complement of $\langle e_1\rangle$, say $\langle e_1+e_2\rangle$? – Frank Science Aug 13 '13 at 0:43
up vote 6 down vote accepted

A very helpful intuition to have in this situation is the notion of Eigenspaces. Certainly, any operator will be invariant over an Eigenspace, which means that diagonalizable matrices won't make for a good counterexample here.

A generally useful matrix to have as a counterexample in many instances (see first answer), including this one, is $$ \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix} $$

A useful habit to form, in any field of mathematics, is to collect (as my professor once put it) a "zoo" of mathematical counterexamples. If you're working in graph theory, keep in mind the Petersen graph. If you're working in topology, keep in mind the topologist's sine curve and the Hawaiian earing. If you're working in linear algebra, keep this matrix in mind.

The more populated and the more diverse your zoo, the better your intuition will be for these and other problems. With surprising frequency, you'll be able to pick the right counterexample out of your zoo and plug it straight in. Other times, you might be able to use one or several counterexamples to build another.

share|cite|improve this answer
I've added the link to go with that matrix. I hope you find this answer useful! – Omnomnomnom Aug 12 '13 at 23:05
Thanks. This one goes into my bag of tricks. :) – Jack Dawkins Aug 13 '13 at 0:11
Another for your bag of tricks then: $$\begin{pmatrix}1&i\\i&i\end{pmatrix}$$ is symmetric but not normal. And you're welcome :) – Omnomnomnom Aug 13 '13 at 0:23
I'm sorry I took this long to accept your answer. I totally forgot. – Jack Dawkins Jun 13 '14 at 19:59
Ha! Not a problem :D – Omnomnomnom Jun 13 '14 at 20:04

Intuitively, you might ask yourself: why should $T$ preserve a complementary subspace of $W$? Think of some basic linear operators you know, and ask if they do this. I thought of the operator which is a projection onto the subspace $W$ along a subspace other than $U$. Then $U$ is not $T$-invariant.

share|cite|improve this answer
Thanks Eric. This helps. – Jack Dawkins Aug 13 '13 at 0:11

It's easier to come up with counter-examples if you understand more about how matrices and linear transformations work, e.g. via Jordan normal form and the special form of complex-valued eigendecompositions of real matrices. E.g. if you understand the meaning of Jordan normal form, then you know that it is possible to not have a full eigen-decomposition, and instead have a matrix that "shifts" some vector to a combination of itself and some other vector, rather than having a full set of eigenvectors each of which is carried to a multiple of itself. This is exactly the kind of counter-example you found: You wrote down a matrix that leaves invariant the $e_1$ subspace but shifts $e_2$ to a combination of $e_1$ and $e_2$ (thus does not leave invariant the $e_2$ subspace). Realizing this kind of linear transformation is possible, and how to construct it, should be more obvious to you once you know more about the fundamental characterizations of matrices and linear transformations and what they mean.

share|cite|improve this answer

Supposing you couldn't come up with a counterexample, here's what you could have done: classify all $T$ and all $W$ for which this is true.

Let's assume that $U$ is $T$-invariant for all subspaces $U$ of $V$ such that $V=W\oplus U$. Now let's fix one such $U$. If $u_1,\dots,u_n$ is a basis for $U$, then $u_1+w,\dots,u_n+w$ is a basis for a subspace $U_w$ such that $V=W\oplus U_w$. Therefore, $U_w$ is $T$-invariant for all $w\in W$ by assumption. In particular, $T(u_i+w)=\sum_{j_i=1}^{n} a_{j_i}u_i +(\sum_{j_i=1}^{n} a_{j_i})w$ for all $1\leq i\leq n$. Since $U$ and $W$ are $T$-invariant, we obtain a common element in $U$ and $V$: $Tu_i-\sum_{j_i=1}^{n} a_{j_i}u_i=(\sum_{j_i=1}^{n} a_{j_i})w-Tw$ which must be zero. Therefore, all $w\in W$ are eigenvectors of $T$.

Can you take this further, as an exercise, and see what properties $T$ must satisfy and ultimately classify all such $T$? Or is this as far as we can go, in general: if the pair consisting of $T$ and $W$ satisfies this property, then every vector in $W$ is an eigenvector of $T$?

I hope this helps!

share|cite|improve this answer

Your Answer


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.