## Monthly Archives: November 2008

### linear algebra questions

Dear Professor,

I’m a second year Algebra student, and I was hoping you could answer a couple of questions.

1.

a. Notation. In the online notes, page 43, you refer to the set $GL_n (k)$. I don’t recall seeing this notation before – what set does this refer to?

b. On page 45 you twice refer to $q(x,y)_t$ seeming to mean a transposed quadratic form. What does this mean? How can one transpose what is essentially a function?

2.

From my notes from Thursday’s lecture, in an example we consider a quadratic form $q$, $C_1$ the standard basis for $R^3$, $C_2$ a different basis, and $M$ the change of basis matrix from $C_1$ to $C_2$, so that $\left[q\right]_{C_1} = B$, $B$ being a 3×3 matrix which we had earlier diagonalised via double operations to obtain the diagonal matrix $D$.
We called the matrix that effected those double operations $P$, so that

$P^TBP = D.$

Then, changing basis from $C_1$ to $C_2$, we wrote:

$\left[q\right]_{C_2} = M^t\left[q\right]_{C_1}M$

Then the following line is what I don’t understand: “Because $C_1$ is the standard basis for $k^3$, the columns of $P$ are exactly elements of the new basis $C_2$” Why is this the case?

Apologies for the slightly involved question, I am more than happy to come and explain the problem in person – I intended to come to the office hour today, but unfortunately forgot my notes… (!)

Thanks
——————————————————————————————

Reply:

1. a. $GL_n(k)$ refers to the set of $n\times n$ invertible matrices with entries in $k$. It actually forms a group under multiplication. Taking $k$ to be $R$ or $C$ give the most basic examples of Lie groups.

b. This is slightly odd notation, but I’ll let it stand. The point is that we usually view vectors in $k^n$ as column vectors. So I wrote

$q(x,y)^t$

to mean “the function $q$ of the column vector $(x,y)^t$.”

2. There are two facts: If the change of basis matrix from $C_1$ to $C_2$ is $M$, then

$\left[q\right]_{C_2} = M^t\left[q\right]_{C_1}M.$

Conversely, if $C_1$ is a basis, then for any invertible matrix $M$,

$M^t\left[q\right]_{C_1}M$

is the matrix of $q$ with respect to the basis $C_2$ with change of basis matrix $M$ from $C_1$. So in our case, $D$ is the matrix of the quadratic form with respect to the basis $C_2$ with the property that the change of basis matrix from $C_1$ to $C_2$ is $P$. However, how does one calculate the change of basis matrix $M$ from the standard basis

$C_1=\{e_1, e_2, e_3\}$

to a basis $C_2=\{b_1, b_2, b_3\}$? The first column of $M$ is given by the coefficients in the expression of $b_1$ as a linear combination of the $e_i$. But these coefficients are nothing but the entries of $b_1$. Thus, $b_1$ is exactly the first column of $M$. Similarly for the other columns. So in our case, when we pose the question of “what is the basis with respect to which $q$ acquires the diagonal form $D$?”, the answer is given by the columns of the change of basis matrix $P$.

Advertisements

### Note on quadratic forms

For those of you who downloaded the note on quadratic forms from the 2201 course website, you should be aware that there is a revision now available. Some misprints were corrected as well as a significant error where the inverse of the transition matrix should have been used.

### Change of basis for linear maps and bilinear forms

dear professor,

when I read the note, I find these problems, hope you do not mind

————————————————————————————
1. in page 40, at the end of 4.1.100 proposition, you wrote “and for any linear map $T:V \rightarrow V$ we have $\left[T\right]_C= M^{-1}\left[T\right]_B M$,” can you explain why? because in 4.1.101, the note said $\left[f\right]_C = M^{t}\left[f\right]_B M$, one is $M^{-1}$ and one is $M^{t}$ which makes me a little confused.

——————————————————————————-
2. in page 40, I think all the appearance of “theorem 5.1” should be changed to “theorem 4.1”. am I right?
——————————————————————————-

thanks very much
———————————————————————————

Reply:

1. The intention of that paragraph was exactly to warn you that the way the matrix of a bilinear form changes is different from the way the matrix of a linear map changes when we go from one basis to another.

2. Yes, it should have been theorem 4.1. Thanks very much for the correction.

### Vector question

Dear Professor.

I want to ask you a maths question. Hope you don’t mind.

What is the difference between

$|C| = \frac {A \cdot B} {|B| },$

and

$C = \frac {A \cdot B} {|B|^2} B$ ?

Where $C$ is the projection of $A$ on $B$.

I understand that the first expression is the magnitude of $C$, which is the length of the projection, am i right?

However, I’m uncertain about the second expression and to what significance it holds. Please enlighten me regarding this.

Thank you very much! Your help is much appreciated.

Take care!

—————————————————————————————-
Reply:

You are right about the length of the projection. Perhaps the best way to see the meaning of the second expression is to write it as:

$C=(A \cdot (\frac{B}{|B|}))\frac{B}{|B|}$

Here is the point: Given any vector $W$, I believe you know how to visualize its projection along any directed line $L$. For simplicity, I will assume that $W$ forms an acute angle with the direction of the line. The projection is then the shadow of $W$ you see when you shine a light on $L$ along a perpendicular angle. It is an elementary exercise with a triangle to see that the length of this projection is exactly

$W\cdot e_L,$

where $e_L$ is a *unit vector*, that is, a vector of length 1, in the direction of $L$. You should think of a unit vector as encoding ‘pure direction.’ So to get a vector of magnitude $r$ in the direction of the unit vector $E$, you can just dilate $E$ by a factor of $r$ to get

$X=rE.$

Conversely, to change any non-zero vector $V$ into the unit vector in the same direction, you divide by its magnitude to get

$\frac{V}{|V|}.$

Note now the key fact that the projection of $W$ itself, not just its magnitude, is given by the formula

$(W\cdot e_L)e_L$

since the projection is clearly in the direction of $e_L$ and this expression gives such a vector with just the right magnitude.

To return to your question, given any two vectors $W$ and $V$, you can project $W$ in the direction of $V$. Now, the unit vector in the direction of $V$ is exactly

$\frac{V}{|V|}.$

So the projection of $W$ in the direction of $V$ is nothing but

$(W\cdot (\frac{V}{|V|}))\frac{V}{|V|}.$

By the way, your question is a very good one. It’s very important to understand the geometric meaning of basic expressions like this. Otherwise, everything degenerates into a bunch of formulas.

### Integral bases

Because my lecture just before the break was rather vague about the algorithm for computing integral bases, I’ve written up a note that explains it in greater detail. Here is a link.