3
$\begingroup$

I've recently seen the example on how a shear $\begin{bmatrix}1&1\\0&1\end{bmatrix}$ is an example of a defective matrix, since it has eigenvalues $1,1$ but only one independent eigenvector $\mathbf{v}_1=(1,0)$.

So in this case, I can see that the algebraic multiplicity is greater than the geometric multiplicity, But what I am wondering is why the system created "two" eigenvalues when only one of them was "actually an eigenvalue" - (It's not even that there were two eigenvectors where each one corresponded $1$, $1$).

Is there any underlying reason why the system came up with two eigenvalues? (If I was to intuitively guess what would happen purely from a geometric perspective, I would have guessed that the characteristic polynomial would just be linear: $\lambda - 1$. Although this isn't possible for a 2x2, I was wondering whether there was any other meaning for the second eigenvalue?)

$\endgroup$
  • $\begingroup$ You’re on the track of something important in your last paragraph: The characteristic polynomial of an order-$n$ square matrix is by definition always of degree $n$, but that’s not the whole story. There’s also the minimal polynomial, which is the least-degree monic polynomial for which $p(A)=0$. A matrix is diagonalizable iff its minimal polynomial factors into distinct linear terms. $\endgroup$ – amd Jul 29 at 23:29
2
$\begingroup$

Well, I would say there is only one eigenvalue: $1$. The point is, we usually say it's "repeated" or that it has "(algebraic) multiplicity $2$". Think about what it means to "repeat" an eigenvalue; under which circumstances do we list it twice, or more? And, when we do, how many times should we be listing it?

You seem to be counting based on the dimension of the eigenspace for the eigenvalue (or equivalently, the maximum number of linearly independent eigenvectors you can come up with). This is known as the geometric multiplicity. And, indeed, the geometric multiplicity of $1$ is $1$ in this case. Note how it does not agree with the exponent of the $\lambda - 1$ factor in the characteristic polynomial.

The algebraic multiplicity counts the dimension of the generalised eigenspace. The generalised eigenspace is given by $$\operatorname{ker}(M - \lambda I)^n$$ where $M$ is an $n \times n$ matrix and $\lambda$ is an eigenvalue. Note how this contains $\operatorname{ker} (M - \lambda I)$ (if $(M - \lambda I)$ sends a vector to $0$, then applying it $n - 1$ more times will still send it to $0$), which is the (usual) eigenspace corresponding to $\lambda$. When $M$ is diagonalisable, this is invariably equal to $\operatorname{ker}(M - \lambda I)$, but when $M$ is defective, this can be larger than the eigenspace.

Now, as it turns out, the generalised eigenspaces always sum to $\Bbb{C}^n$, and as a consequence, we can always form a basis of generalised eigenvectors. There's a particularly nice class of such bases called Jordan bases; these are the next best things we can find to bases of eigenvectors. Instead of diagonalising a matrix, they turn it into Jordan Normal Form, an excellent consolation prize when a diagonal representation is denied to us. Jordan normal forms exist for every matrix, unlike diagonal forms!

The algebraic multiplicities also correspond to the exponents of the corresponding factors in the characteristic polynomials. In fact, some define the characteristic polynomial by this characteristic, and its determinant representation becomes a theorem.


In the case of the $2 \times 2$ matrix presented, the eigenspace, $\operatorname{ker} (M - I)$ is simply $\operatorname{span}\{(1, 0)\}$. However, if we compute $$\operatorname{ker} \left(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\right)^2 = \operatorname{ker} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}^2 = \operatorname{ker} \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = \Bbb{C}^2,$$ we see that the generalised eigenspace is $2$-dimensional, and the algebraic multiplicity is $2$.

$\endgroup$
  • $\begingroup$ Thank you for your reply it is very clearly explained at a level I could understand, this is really quite insightful. I'm wondering, is this generalised eigenspace analogous to repeated factors like $(\lambda - 1)^3$ and how they are roots to a function and it's derivative and so on? - I haven't met them before $\endgroup$ – user523384 Jul 29 at 9:06
  • $\begingroup$ These are decently-sized questions. Developing this connection properly takes approximately a whole university course. I suggest you read Axler's Linear Algebra Done Right, or at least the chapter on generalised eigenvectors and Jordan Normal Form. Just be warned: he has a non-standard take on determinants, and defines them at the end of the book. So, if you want to know the (entirely non-trivial) connection between the characteristic polynomial as defined by dimensions of generalised eigenspaces, and the formula using cofactors, prepare to do some reading. $\endgroup$ – Theo Bendit Jul 29 at 10:31
1
$\begingroup$

I don't know if this qualifies as intuition, but in your case note that $$ \begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}+\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix} $$

One of these is a proper diagonal matrix, while the other is a shift-matrix. Now, it is a theorem that for any nil-potent matrix $A$, i.e. a matrix such that $A^n=0$ for some $n$, there exists some basis $\mathcal{B}=\{b_1,...,b_n\}$ such that the matrix in this basis has the form of a shift matrix (i.e. $A b_i= b_{i+1}$ or $A b_i=0$ for any $i$).

There is another theorem stating that any matrix is the sum of a nil-potent matrix and one that is diagonalisable (as long as your scalars belong to an algebraically closed field, such as $\mathbb{C}$). Hence, take your general $M=D+A,$ where $D$ is diagonalisable and $A$ is nil-potent. Then, the above observations allows us to find some basis such that $D$ is diagonal and another basis such that $A$ is a shift-matrix.

However, the magic of the Jordan Normal Form is that this can be done simultaneously. I.e., there exists a single basis with respect to which $D$ is diagonal and $A$ is a shift-matrix. Thus, the obstruction to the geometric multiplicity of every eigenvalue corresponding to their algebraic multiplicity is exactly this shift-matrix $A$. If $A=0,$ then $D$ is diagonalisable and, of course, vice versa.

In your case, you have something that is almost an eigenvector, namely $(0,1),$ but instead of just producing a scalar multiple of itself, it produces itself and an honest-to-god eigenvector, $(1,0)$. These are called generalised eigenvectors.

So to sum up: Geometrically speaking, you don't have as many actual eigenvectors as you would like because some shift is happening.

$\endgroup$
  • $\begingroup$ I upvoted soley because of your second last paragraph, but thank you, already reading helped this make more sense to me, I'm about to learn about the Jordan Normal Form soon so this will definitely be very useful when I do it, Thanks! $\endgroup$ – user523384 Jul 29 at 9:11
1
$\begingroup$

At its core, this comes from the difference between the zero matrix and a nilpotent matrix. A nilpotent $n\times n$ matrix $A$ has characteristic polynomial $x^n$, so it will have only $0$ as an eigenvalue, with algebraic multiplicity $n$. However, its kernel might not be the whole space, and thus the geometric multiplicity might be lower. (The standard extreme case example is a matrix with only $1$ along the superdiagonal, and $0$ otherwise. Then the eigenvalue $0$ has geometric multiplicity $1$.)

Given a nilpotent matrix $A$, and any (compatible) vector $x$, for some natural number $k$ we have $$ A^kx = 0 $$ So if we instead of looking at the kernel of $A$ looked at the combined kernels of any (positive) power of $A$, then you would get the whole space. This is the idea behind generalized eigenvectors.

Given a matrix $B$, the non-zero vector $x$ is an eigenvector of $B$ with eigenvalue $\lambda$ if $$ Bx = \lambda x\\ (B-\lambda I)x = 0 $$ Along the same lines, $x$ is called a generalized eigenvector of $B$ is we have $$ (B-\lambda I)^kx = 0 $$ for some natural $k$. For each eigenvalue $\lambda$, you get a corresponding subspace where $B-\lambda I$ acts like a nilpotent matrix. Its dimension is equal to the algebraic multiplicity of $\lambda$ (and thus the algebraic multiplicity and the generalized geometric multiplicity agree). If the algebraic multiplicity and the geometric multiplicity of $\lambda$ agree, then $B-\lambda I$ acts as the zero matrix on this subspace.

If you use this to "diagonalize" $B$ as well as you can, you get the so-called Jordan normal form (with $\lambda$ along the diagonal, and $1$'s along the superdiagonal at suitable places).

In your case, the matrix is already in Jordan normal form, and the whole plane (minus the origin) are generalized eigenvectors with eigenvalue $1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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