Remember that we've already proven the following, for any real symmetric $n\times n$ matrix $M$: (i) Each eigenvalue of $M$ is real. (ii) Each eigenvector can be chosen to be real. (iii) Eigenvectors with different eigenvalues are orthogonal.

(b) Let $A$ be a real antisymmetric $n\times n$ matrix. Prove that each eigenvalue of $A^{2}$ is real and is less than or equal to zero.

$(A^2)^T = (A^T)^2 = (-A)^2 = A^2$, so $A^2$ is real symmetric. By virtue of (i) above, the eigenvalues of $A^2$ must be real.

$1$. How would you determine to prove that $A^2$ is symmetric, so that you can benefit from (i) ?

Let $A^2v = \color{orangered}{ k \; \mathbf{ v } }$, where k is a scalar. By (ii) above, hypothesise that $\mathbf{ v }$ is real. Then $\begin{align} k \mathbf{ v^Tv } & = v^T \; \color{orangered}{ k \; \mathbf{ v } } = v^T \; \color{forestgreen}{ A^2 }v = v^T \color{forestgreen}{ AA } v = v^T\color{forestgreen}{ (-A^T)A }v \\ & = -(Av)^T(Av) < 0 \end{align}$.

$2.$ The question asks us to prove $k < 0$, but what's the proof strategy? The trick looks like to consider $k \mathbf{ v^Tv } $, but how would you determine/divine/previse this?
I remember $\langle v,v \rangle := v^Tv \ge 0$.

$3.$ I'm not asking about the algebra itself, but what's the strategy behind it here? The last few steps feel too clever/guileful?

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    $\begingroup$ I believe that the part that says Av=kv is erroneous. It should have said "let A^2v=kv", as the assumption is the k is an eigenvalue of A^2. $\endgroup$ – Meni Rosenfeld May 26 '14 at 20:08
  • $\begingroup$ @MeniRosenfeld Thank you very much! I think you're right. I've just emended it. $\endgroup$ – Greek - Area 51 Proposal May 26 '14 at 20:18
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    $\begingroup$ Notice that if $k$ is an eigenvalue of $A$, with eigenvector $v$, then $Av=kv$, thus $A^2v=Akv=kAv=k^2v$. And eigenvalues of an antisymmetric matrix are pure imaginary. Regarding the logic behind your exercise, you may have a look at quadratic forms. $\endgroup$ – Jean-Claude Arbaut May 26 '14 at 20:25
  • $\begingroup$ It seems to me that in $1.$ there should be a $\leq$ in the last inequality. Because an eigenvector $v$ of $A^2$ is nonzero, you can divide the inequality $k v^t v\leq 0$ by $v^t v$ (which is positive) and get $k\leq 0$. $\endgroup$ – Peter Franek May 31 '14 at 7:27

So you are given a matrix with a special property and are asked how to go about finding an inequality of the eigenvalue of this matrix. What is the natural idea?

First, note that most of elementary linear algebra goes through for arbitrary fields. This includes finite fields where there exists no ordering at all. So in order to prove an inequality one has to take advantage of something that is special to $\mathbb{R}$ or $\mathbb{C}$.

This is the fact that any finite dimensional vector space, say $V$, over $\mathbb{R}$ or $\mathbb{C}$ is an inner product space. If you look at the axioms of an inner product space, you can note that the only one involving an inequality is positive definiteness (that is $x \cdot x \geq 0$ for all $x \in V$). You may know of other inequalities that hold true in inner product spaces, say Cauchy-Schwarz and triangle inequality, but these follow from positive definiteness.

Now the primary object in your question is an eigenvalue, $k$ which in its definition you will find an associated eigenvector, $v$.

With this in mind, we want to see if we can somehow apply positive definiteness to your problem. Now we need the particular hypothesis from your problem, i.e. that $A = - A^T$. A lightbulb should immediately go off since $A^T$ is the adjoint (for a real matrix the adjoint is the same as the transpose, but the important property is that $Av \cdot Av = A^T A v \cdot v$) of $A$, a special matrix defined in terms of the inner product! We know $\langle Av , Av \rangle = \langle A^T A v , v \rangle$. From here the problem is easy to finish, using that $A^T = -A$ and $A^2v = kv$.

Let's see another example of the similar idea. Let $V$ be a vector space over $\mathbb{R}$ or $\mathbb{C}$ and let $A: V \to V$ be a linear transformation. We want to show if there is a vector $x \in V$ such that $A^* Ax = 0$, then $A x = 0$, where $A^*$ is the adjoint of $A$.

Again we consider $\langle Ax , Ax \rangle = \langle A^* A x , x \rangle$ and we can see this is $0$ since $A^* A x = 0$. Thus $Ax \cdot Ax = 0$ and so it must be that $Ax = 0$.

  • $\begingroup$ Thanks, but I don't see how " the problem is easy to finish." I calculate $<Av, Av> = (Av)^T A \quad v = v^TA^TA \quad v = <A^TAv, v>$. Then? Also, what happened to k? $\endgroup$ – Greek - Area 51 Proposal May 28 '14 at 13:10
  • $\begingroup$ Will you please to respond in your answer, and not as comments? $\endgroup$ – Greek - Area 51 Proposal May 28 '14 at 13:10
  • $\begingroup$ Moreover, why did you discuss the adjoint of A? Here we're working with the Hermitian? $\endgroup$ – Greek - Area 51 Proposal May 28 '14 at 13:11

I do not understand if you want just a hint for the proof or the proof itself. I am giving you both (in the right order :) ).

The hint:

Spectral theorem.

The proof:

The antisimmetry is conserved through similarity, i.e. if A is antisimmetric, so is G A G^{-1} for any non singular matrix G.

In addition, a real antisimmetric matrix is normal (i.e. it commutes with its adjoint) and hence diagonalizable in complex field. The former observation points out that the diagonalized matrix is antisimmetric too, and hence it's eigenvalues are pure immaginary and the eigenvalues of A^2 are real numbers, less than or equal to zero.

I haven't read very well, but it seems to me that english wikipedia furnish a similar proof: http://en.wikipedia.org/wiki/Skew-symmetric_matrix#Coordinate-free



Sorry, I think I have misunderstood your question. You want to know how to prove the theorem in a way coherent with the rest of the exercise, i.e. using i),ii) and iii), am I right now?

If so, as George Shakan wrote before me you could do this way:

$(,)$ be the standard product in $\mathbb{R}$^n. $A^2$ is real, simmetric (You showed this part). Recall that the transpose is also the adjoint with the standard product. Be $v$ an eigenvector, by i) and ii) it is real and so is the relative eigenvalue $k$, now you have $k = (v,A A v) = (A^T v, A v) = - (A v, A v) < 0$. Note that you need the eigenvalue/vector to be real so that $(v, A A v)$ make sense.

  • $\begingroup$ Thanks, but how does this answer my questions? $\endgroup$ – Greek - Area 51 Proposal May 28 '14 at 17:26

Here's how I see the argument:

As you observe, for any vector $v$, $\langle v, v\rangle \ge 0$. Moreover, $\langle v, v \rangle = 0$ if and only if $v=0$. Now, let $v$ be an eigenvector of $A^2$, and let $k$ be the corresponding eigenvalue. Then $v \not= 0$ by the definition of an eigenvector, so $\langle v, v \rangle > 0$. We also have that $v$ and $k$ are both real. Let's see what we can do with that.

Since $v$ is an eigenvector of $A^2$, and by the linearity of the inner product, we obtain: $$\langle A^2v, v \rangle = \langle kv, v\rangle = k \langle v, v \rangle.$$

But also, $$\langle A^2v, v \rangle = (A^2v)^T v = v^T (A^2)^T v = v^T (A^T)^2 v$$ $$= v^T A^T A^T v = (v^T A^T) (-Av) = (Av)^T(-Av) = \langle Av, -Av\rangle$$ $$= - \langle Av, Av\rangle.$$

From these two expressions, we see that $k\langle v, v \rangle = -\langle Av, Av \rangle$. The right hand side of this inequality is less than or equal to zero, so $k\langle v, v\rangle \le 0$. Since the inner product $\langle v, v \rangle > 0$, it follows that $k \le 0$.


1) You want to prove that every eigenvalue of $A^2$ is real and negative. Since every eigenvalue of a real symmetric linear operator is real, proving that $A^2$ is symmetric would get you half of what you need.

2) I think the stated argument is unclear. It's much clearer to start with $A^2(v) \cdot v$ instead, as this would more clearly let one assess whether $A^2$ is positive- or negative-definite. In fact, it's much easier and faster to prove directly that $A^2$ is negative definite:

$$A^2(b) \cdot b = A(b) \cdot A^T(b) = -A(b) \cdot A(b) < 0$$

But maybe you aren't allowed to invoke the properties of a negative-definite operator. You can still use the above argument to suggest that the eigenvalue $k$ must necessarily be negative.

To the extent that the proof is convenient, or that it is unclear how one would attack the problem, I cannot comment much. Here, we find the sign of $k$ by showing that $k|v|^2 \leq 0$ and knowing that $|v|^2 \geq 0$. Using the known signs of several individual quantities and the overall sign of a quantity to solve for the sign of an unknown? I'm sure this was done in basic algebra also.

Comment responses:

The definition of the transpose is the map $A^T$ such that

$$A(b) \cdot c = A^T(c) \cdot b$$

This is a general, and very useful, definition. That works even when talking about complex vector spaces (where instead of the transpose, we call this kind of map the adjoint instead).

In this problem, we just do a little trick like so:

$$A^2(b) \cdot b = A[A(b)] \cdot b = A^T(b) \cdot A(b)$$

Finally, remember that the inner product is positive definite: the dot product of a vector with itself is always greater than zero (unless it's the zero vector). Here, we're considering the dot product of $A(b)$ with itself, so that's positive, and the minus sign incurred by swapping from $A^T$ to $A$ ensures that the quantity is overall negative.

Remember that inner products are commutative. $x \cdot y = y \cdot x$. Similarly, $A(b) \cdot c = c \cdot A(b)$. In this case, converting to matrix notation:

$$A(b) \cdot c = (Ab)^T c = b^T A^T c = b^T (A^T c) = b \cdot A^T(c) = A^T(c) \cdot b$$

Thus, there is no contradiction.

My notation is purposefully avoiding any mention that $A$ can be represented by some matrix, or that inner products can be written in terms of matrix multiplication of row and column vectors. None of the important results of linear algebra rely upon representing vectors and linear maps with row/column vectors and matrices. The latter are simply a means to performing computations.

Moreover, when working in non-Euclidean spaces, one would have to make some modifications: first, the notion of transpose being important no longer applies (obvious when considered in context of complex spaces, but it's true even in something as simple as a Minkowski spacetime); second, inner products rely upon some non-identity matrix to sit between the row and column vectors, apparently giving the Euclidean inner product a privileged character, a privileged relationship to the identity matrix that is merely an artifact of how we define matrix multiplication.

  • $\begingroup$ Thanks. I don't perceive your notation: Why $A^2(b) \cdot b = A(b) \cdot A^T(b)$ ? And why $ -A(b) \cdot A(b) < 0 $ ? $\endgroup$ – Greek - Area 51 Proposal May 28 '14 at 13:01
  • $\begingroup$ I have done so, as you requested. $\endgroup$ – Muphrid May 28 '14 at 14:07
  • $\begingroup$ THank you. I still don't see $A(b) \cdot c = A^T(c) \cdot b$? My textbok says $<x,y> = x \cdot y = x^Ty$ ? SO here, your $Ab \cdot c = (Ab)^Tc = b^TA^T c$, but this differs from your supplementary? $\endgroup$ – Greek - Area 51 Proposal May 28 '14 at 17:27
  • $\begingroup$ Will you please to respond in your answer, and not as comments? $\endgroup$ – Greek - Area 51 Proposal May 28 '14 at 17:28
  • $\begingroup$ I've added a section on commutativity of inner products and general remarks about the use of matrices in linear algebra. $\endgroup$ – Muphrid May 28 '14 at 18:55

Symmetric matrix $B$ has non positive egenvalues iff for any non zero vector $v$ the quadratic form $v^TBv\le 0$. In our case $B=A^2=A(-A^T)$ so $v^TBv=-v^TAA^Tv=-||A^Tv||\le 0$.


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