Abstract Nonsense

Crushing one theorem at a time

Halmos Section 23: Bilinear Forms

Point of post: This post will document the solutions to Halmos’s section 23 on Bilinear forms, the content of which is explained (not quoted) here. Some interesting results in this post is the, very very perfunctory, discussion of quadratic forms and a how a symmetric bilinear form on a field of characteristic greater than two is completely determined by it’s associated quadratic form.

Remark: As usual, I use somewhat different notation than Halmos. To see this, and examples, etc. see the above link.



a) If B\in\text{Bil}\left(\mathbb{R}^n,\mathbb{R}^n\right), then there exists scalars \gamma_{i,j},\text{ }i,j\in[n] such that

\displaystyle B\left(\sum_{i=1}^{m}\alpha_i x_i,\sum_{j=1}^{n}\beta_j y_j\right)=\sum_{i=1}^{m}\sum_{j=1}^{n}\alpha_i\beta_j\gamma_{i,j}

b) If \varphi\in\text{Hom}\left(\text{Bil}\left(\mathbb{R}^n,\mathbb{R}^n\right),\mathbb{R}\right) then there exists scalars \gamma_{i,j}\text{ }i,j\in[n] such that

\displaystyle \varphi=\sum_{i=1}^{m}\sum_{j=1}^{n}\gamma_{i,j}\eta_{i,j}


a) This follows immediately from my discussion in the above link by taking \gamma_{i,j}=B(x_i,y_j)

b) This is true in a much more general setting as we shall see in the next section, but for now it suffices to note that for any B\in\text{Bil}\left(\mathbb{R}^n,\mathbb{R}^n\right) we have that

\displaystyle B=\sum_{i=1}^{m}\sum_{j=1}^{n}B(x_i,y_j)B_{i,j}

Thus, note that if \varphi is a linear functional on \text{Bil}\left(\mathbb{R}^n,\mathbb{R}^n\right) that the above implies that

\displaystyle \varphi\left(B\right)=\sum_{i=1}^{m}\sum_{j=1}^{n}B(x_i,y_j)\varphi\left(B_{i,j}\right)

and clearly this representation is unique. So, taking B(x_i,y_j)=\gamma_{i,j} and \varphi\left(B_{i,j}\right)=\eta_{i,j} gives the desired result.


Problem: A bilinear form \mathscr{U}\boxplus\mathscr{V} is degenerate if, as a function of one of its two arguments, it vanishes identically for some non-zero value of its other argument, otherwise it is non-degenerate

a) Give an example of a degenerate linear form (not identically zero) on \mathbb{C}^2\boxplus\mathbb{C}^2

b) Give an example of a non-degenerate bilinear form on \mathbb{C}^2\boxplus\mathbb{C}^2.

Proof: Define

f:\mathbb{C}^2\boxplus\mathbb{C}^2\to\mathbb{C}:\left(\begin{bmatrix}z_1 \\ z_2\end{bmatrix},\begin{bmatrix}z_3 \\ z_4\end{bmatrix}\right)\mapsto (z_1+z_2)(z_3+z_4)

Evidently f is linear (this really isn’t that hard to check), but notice that

f\left(\begin{bmatrix}1\\ -1\end{bmatrix},\begin{bmatrix}z_3\\ z_4\end{bmatrix}\right)=0,\text{ }\forall \begin{bmatrix}z_3 \\ z_4\end{bmatrix}\mathbb{C}^2

b) Consider \det (as defined in the last post). Then, \det is a bilinear form on \mathbb{C}^2\boxplus\mathbb{C}^2 but notice that

\det\left(\begin{bmatrix}z_1\\ z_2\end{bmatrix},\begin{bmatrix} 0\\ 1\end{bmatrix}\right)=0\implies z_1=0


\det\left(\begin{bmatrix}z_1 \\ z_2\end{bmatrix},\begin{bmatrix} 1\\ 0\end{bmatrix}\right)=0\implies z_2=0

Thus, if we fixed the first entry of \det, then the only way it can be zero for all second entries is if it’s first entry is zero. Considering that interchanging the columns (switching the first and second entries) only changed the sign of \det the conclusion follows.


Problem: If B\in\text{Bil}\left(\mathscr{U},\mathscr{V}\right), if y_0\in\mathscr{V} and if \varphi is defined by \varphi(x)=B(x,y_0), prove that \varphi\in\text{Hom}\left(\mathscr{U},F\right). Is it true that if B is non-degenerate, then every linear functional on \mathscr{U} can be obtained this way?

Proof: Evidently such a \varphi is a linear functional since

\varphi(\alpha x_1+\beta x_2)=B(\alpha x_1+\beta x_2,y_0)=\alpha B(x_1,y_0)+\beta B(x_2,y_0)=\alpha\varphi(x_1)+\beta\varphi(x_2)

Now, consider the map

f_B:\mathscr{V}\to\text{Hom}\left(\mathscr{U},F\right):y_0\mapsto B(x,y_0)

clearly this is linear since

f_B(\alpha y_1+\beta y_2)= B(x,\alpha y_1+\beta y_2)=\alpha B(x,y_1)+\beta B(x,y_2)=\alpha f_B(y_1)+\beta f_B(y_2)

Also note that if B is non-degenerate then

B(x,y_1)=B(x,y_2)\implies B(x,y_1-y_2)=\bold{0}

and by assumption this implies that y_1-y_2=0. Thus, f is injective. It follows that

\mathscr{V}\cong f_B\left(\mathscr{V}\right)\leqslant\mathscr{U}

but if \dim_F \mathscr{U}=\dim_F\text{Hom}\left(\mathscr{U},F\right)>\dim_F\mathscr{V} this function cannot be an isomorphism, and thus cannot be surjective. For a concrete example consider \mathscr{U} to be \mathbb{C} over \mathbb{R} and \mathscr{V} to \mathbb{R} over itself. Then, by the above analysis any non-degenerate function B cannot induce a surjection f_B. But, let’s show that for a specific example. Consider

B:\mathbb{C}\boxplus\mathbb{R}\to\mathbb{R}:(\alpha+\beta i,\gamma)\mapsto \gamma(\alpha+\beta)

Then, we note that

f_B(\gamma_0)(\alpha+\beta i)=\gamma_0(\alpha+\beta)

but, clearly f(\alpha+\beta i)=\alpha is a linear functional and not of the above form.


Problem: Suppose that for each p,q\in\mathcal{P}_n, the function B is defined by

a) \displaystyle B(p,q)=\int_0^1 p(x)q(x)dx

b) B(p,q)=p(1)+q(1)

c) B(p,q)=p(1)q(1)

d) B(p,q)=p(1)q'(1)


a) This is evidently.

\displaystyle \begin{aligned}B(\alpha p_1+\beta p_2, q)&= \int_0^1 (\alpha p_1(x)+\beta p_2(x))q(x)dx\\ &=\alpha\int_0^1 p_1(x)q(x)+\beta\int_0^1 p_2(x)q(x)dx\\ &=\alpha B(p_1,q)+\beta B(p_2,q)\end{aligned}

Since B(p,q)=B(q,p) the rest follows by symmetry.

b) This is not since B(0(x),1(x))=1

c) This is since

\begin{aligned}B(\alpha p_1+\beta p_2,q)&=(\alpha p_1(1)+\beta p_2(1))q(1)\\ &=\alpha p(1)q(1)+\beta p(1)q(1)\\ &=\alpha B(p_1,q)+\beta B(p_2,q)\end{aligned}

And the other direction follows by symmetry

d) This is, the case that B is linear in the first entry is trivial (done above basically) and the second follows since

\displaystyle \begin{aligned}B(p,\alpha q_1+\beta q_2)&=p(1)\left(\alpha q_1+\beta q_2\right)'\mid_{x=1}\\ &= p(1)\left(\alpha q_1'(1)+\beta q_2'(1)\right)\\ &= \alpha p(1)q_1'(1)+beta p(1)q_2'(1)\\ &= \alpha B(p,q_1)+\beta B(p,q_2)\end{aligned}

a) This is non-degenerate since if one fixed p_0(x)\in\mathcal{P}_n then if B(p_0,q)=0 for all q, then in particular

\displaystyle \int_0^1 p(x)^2dx=0 \implies p=0

c) This is degenerate since B(x-1,q)=0 for all q\in\mathcal{P}_n

d) Similarly noting that B(p,1(x))=0 for all p\in\mathcal{P}_n we see that B is degenerate.


Problem: Does there exist a vector space \mathscr{V} and a bilinear form B\in\text{Bil}\left(\mathscr{V},\mathscr{V}\right) such that B\ne\bold{0} but B(x,x)=0 for all x\in\mathscr{V}?

Proof: Yes. Let \mathscr{V}=\mathbb{C}^2 and take B=\det. Then clearly this is not the zero form since

\displaystyle \det\left(\begin{bmatrix}1\\ 0 \end{bmatrix},\begin{bmatrix} 0\\ 1 \end{bmatrix}\right)=1


\displaystyle \det\left(\begin{bmatrix} z_1 \\ z_2 \end{bmatrix},\begin{bmatrix} z_1 \\ z_2 \end{bmatrix}\right)=z_1z_2-z_2z_1=0



a) A bilinear form on \mathscr{V}\boxplus\mathscr{V} is symmetric if B(x,y)=B(y,x) for all x,y\in\mathscr{V}. A quadratic form on \mathscr{V} is a function q:\mathscr{V}\to F by taking q(x)=B(x,x) for some B\in\text{Bil}\left(\mathscr{V},\mathscr{V}\right). Prove that if \text{char }F\ne2, then every symmetric bilinear form is uniquely determined by the corresponding quadratic form. What happens if \text{char }F=2?

b) Can a non-symmetric bilinear form define the same quadratic linear form as a symmetric one?


a) As is common in field theoretic literature denote 1_F+1_F=2_F . Note then that since \text{char }F\ne2 we (and it evidently does not equal one, by the assumption on fields that 1_F\ne0_F) see that 2_F\ne0_F and so 2_F has an inverse \frac{1}{2_F}. But, notice that (remembering that we assumed that B is symmetric)

\begin{aligned}q(x+y) &=B(x+y,x+y)\\ &= B(x,x+y)+B(y,x+y)\\ &= B(x,x)+B(x,y)+B(y,x)+B(y,y)\\ &= B(x,x)+2_F B(x,y)+B(y,y)\\ &= 2_F B(x,x)+q(x)+q(y)\end{aligned}\

and thus from prior discussion we may conclude that

\displaystyle B(x,y)=\frac{1}{2_F}\left(q(x+y)-q(x)-q(y)\right)

Thus, if B'\in\text{Bil}\left(\mathscr{U},\mathscr{V}\right) was such that B'(x,x)=q'(x)=q(x) and symmetric we note that for every (x,y)\in\mathscr{U}\boxplus\mathscr{V} we have that

\displaystyle \begin{aligned}B'(x,y) &= \frac{1}{2_F}\left(q'(x+y)-q'(x)-q'(y)\right)\\ &=\frac{1}{2_F}\left(q(x+y)-q(x)-q(y)\right)\\ &= B(x,y)\end{aligned}

Evidently a problem can occur if \text{char }F=2 since 2_F will not have a multiplicative inverse.

b) Yes. What about the zero form \bold{0}:\mathbb{C}^2\boxplus\mathbb{C}^2\to\mathbb{C} and \det:\mathbb{C}^2\boxplus\mathbb{C}^2\to\mathbb{C}? The first is clearly symmetric since

\displaystyle \bold{0}\left(\begin{bmatrix}z_1 \\ z_2 \end{bmatrix},\begin{bmatrix} z_3 \\ z_4 \end{bmatrix}\right)=0=\bold{0}\left(\begin{bmatrix}z_3 \\ z_4 \end{bmatrix},\begin{bmatrix} z_1 \\ z_2 \end{bmatrix}\right)

for all \displaystyle \left(\begin{bmatrix}z_1 \\ z_2 \end{bmatrix},\begin{bmatrix} z_3 \\ z_4 \end{bmatrix}\right)\in\mathbb{C}^2\boxplus\mathbb{C}^2. Also, \det is non-symmetric since

\det\left(\begin{bmatrix}1\\ 0 \end{bmatrix},\begin{bmatrix} 0 \\ 1\end{bmatrix}\right)=1\ne -1=\det\left(\begin{bmatrix} 0 \\ 1\end{bmatrix},\begin{bmatrix} 1 \\ 0\end{bmatrix}\right)

That said, if q_1,q_2 are the quadratic forms for \bold{0},\det respectively then we see (as was shown earlier)

q_1\left(\begin{bmatrix} z_1 \\ z_2 \end{bmatrix}\right)=0=q_2\left(\begin{bmatrix} z_1 \\ z_2 \end{bmatrix}\right)

for any \begin{bmatrix}z_1 \\ z_2 \end{bmatrix}\in\mathbb{C}^2.


1. Halmos, Paul R. Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print.


October 27, 2010 - Posted by | Fun Problems, Halmos, Linear Algebra | , , , , ,

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