## 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
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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 $P$.