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There are things in linear algebra that I would like to better understand, from an intuitive point of view. For example, matrices are entities that may be use to transform a domain into another, by means of rotation, shifting, stretching and so on. My first question is this: how much information do I need to know about a matrix to have an idea about what the matrix is doing? Would it be possible to get some rough idea about this without much information about its spectral properties, and say "Well just by looking at the structure of this matrix, and the numbers in it, I can guess this and that...".

The second question is if anyone knows fast method to compute eigenvalues/eigenvector in your head. I know people who can do this for any 2-be-2 matrix, but I don't know which methods are they using.

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For a 2 x 2 matrix it is easy to calculate the determinant. So solving $det(A - \lambda{}I) = 0$ isn't hard and can be (relatively) easily done in your head. Also keep in mind that diagonal matrices hold eigenvalues on their diagonal. –  user12205 Jun 27 '11 at 23:01
    
that's correct. But I've seen people who can do it for any 2-by-2 matrix in like two seconds (literally), so I was wondering if it's just a matter of practice. –  ACAC Jun 27 '11 at 23:07
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If $M=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ then $\det M = ad-bc$ and $\mathrm{Tr} M = a + d$. Therefore $\lambda_1 \lambda_2 = ad - bc$ and $\lambda_1+\lambda_2 = a+d$. –  Yuval Filmus Jun 27 '11 at 23:21
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Dear Bob, have you read Sheldon Axler's textbook Linear Algebra Done Right? I think this is a fantastic book since it presents linear algebra in a very intuitive manner and avoids the use of determinants for the most part (except at the end where determinants are briefly discussed for the sake of completeness). For example, the characteristic polynomial is defined without using determinants and this leads to a simple proof of the Cayley-Hamilton theorem. I wonder how many students can explain intuitively what "$\text{det}(A-\lambda I)=0$" means; it is a mess. –  Amitesh Datta Jun 28 '11 at 1:17
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If $c=0$ in Yuval's matrix $M$, then the eigenvalues are $a$ and $d$. (The eigenvalues of an upper triangular matrix are precisely the entries on the diagonal of the matrix.) –  Amitesh Datta Jun 28 '11 at 2:05

3 Answers 3

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how much information do I need to know about a matrix to have an idea about what the matrix is doing?

When I took Linear Algebra, it was really helpful to visualize the geometry of course! This is a really worthwhile endeavor in any branch of mathematics. When you're given a matrix, keep in mind always that you're looking at linear transformation from one vector space to another. The matrix can be thought of as column (row) vectors, depending on your point of view. Rather than try (in vain) to describe all of the geometry contained within a matrix, I will point you in the direction to begin:

If you haven't yet, please look at MIT's open courseware Linear Algebra video lectures by Dr. Gilbert Strang. His lectures are very good. I haven't studied his book, but I've studied from other books (both cheap and very good: Shilov's Linear Algebra and Axler's Linear Algebra Done Right; buy/rent/obtain them somehow).

The point of eigen(german for characteristic)values is that these vectors remain the same(up to a scaling factor, $\lambda$, under the linear transformation. Take a linear transformation that rotates by 180 degrees in $\mathbb{R}^2$. The vector $(1,0)$ is mapped to $(-1,0)$. This is peculiar: it seems we have only multiplied this vector by a $\lambda = -1$. You can probably guess this is going to be an eigenvalue (actually two). $(0,1)$ is similarly mapped to $(0,-1)$. And in general a vector $(a,b)$ is mapped to $(-a,-b)$. This computation is much easier when the matrix is in its diagonalized form (two -1's on the diagonals).

The second question is if anyone knows fast method to compute eigenvalues/eigenvector in your head. I know people who can do this for any 2-be-2 matrix, but I don't know which methods are they using.

You can try $\det{(A-\lambda I)}=0$. This determinant is not hard, and how the polynomial splits should be recognizable. There are other methods, of course. Maybe you could provide an example matrix? (see Axler for Linear Algebra with little mention of the determinant...) Hopefully you remember that if you have a diagonal matrix $A$, the eigenvalues are on the diagonal :-).

I hope this helped a bit.

Edit: Ohh! I forgot! What Yuval said is very good for quickly finding the product and sum of the eigenvalues.

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So, you want to be able to just look at a $2\times2$ matrix, and see its geometric effect on the plane. I don't think you can always do this, but here are some special cases. I'll assume real entries.

  1. If one column is just a scalar multiple of the other, then the matrix flattens the plane into a line. For example, if $A=\pmatrix{4&6\cr6&9\cr}$ so the first column is two-thirds the second, and the matrix takes everything to the line through the origin and $(4,6)$ (the first column).

  2. If $A=\pmatrix{a&-b\cr b&a\cr}$ then $A$ is a dilation by the factor $\sqrt{a^2+b^2}$ and a rotation. Write $A={1\over\sqrt{a^2+b^2}}\pmatrix{\cos\theta&-\sin\theta\cr\sin\theta&\cos\theta\cr}$ where $\tan\theta=b/a$, and you'll see the rotation is counterclockwise through $\theta$.

  3. If $A=\pmatrix{1&a\cr0&1\cr}$ then $A$ fixes $(1,0)$ but moves $(0,1)$ horizontally to $(a,1)$, so it is a shear.

  4. If $A=\pmatrix{r&0\cr0&s\cr}$ then it stretches the $x$-direction by a factor of $r$ (if $r\gt1$; shrinks by a factor of $r$, if $0\lt r\lt1$), and the $y$-direction by a factor of $s$. If $r$ (or $s$) is negative, then it also does a reflection in the $y$ (or $x$) axis.

Well, that should get you started.

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You can use the singular value decomposition to learn the geometry of a linear transformation. In short, SVD says that every linear transformation maps the unit sphere in the domain to an ellipsoid in a subspace of the codomain.

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