LAC Exercises (with solutions)

Section 3.3 Exercises (with solutions)

Exercises Exercises

1.

Let \(A\in M_n(\R)\) which is invertible. Show that the columns of \(A\) form a basis for \(\R^n.\)

Solution.

Since \(A\) is invertible, we know that we can find its inverse by row reducing the augmented matrix

\begin{equation*} [A|I_n] \mapsto [I_n|A^{-1}]. \end{equation*}

In particular, this says that the RREF form of \(A\) is \(I_n.\)

One way to finish is that the information above says that \(Ax=0\) has only the trivial solution, which means by Observation 1.3.2 that the \(n\) columns of \(A\) are linearly independent. Since there are \(n = \dim \R^n\) of them, by Theorem 3.1.6, they must be a basis.

Another approach is that the linear map \(T:\R^n\to \R^n\) given by \(T(x) = Ax\) is an isomorphism with the inverse map being given \(x\mapsto A^{-1}x\text{.}\) In particular, \(T\) is surjective and its image is the column space of \(A.\) That means that the \(n\) columns of \(A\) span all of \(\R^n,\) and hence must be a basis again by Theorem 3.1.6.

2.

Consider the vector space \(M_2(\R)\) of all \(2\times 2\) matrices with real entries. Let’s consider a number of subspaces and their bases. Let \(\cE=\{E_{11},E_{12},E_{21},E_{22}\} = \{ [\begin{smallmatrix} 1\amp0\\0\amp0 \end{smallmatrix}], [\begin{smallmatrix} 0\amp1\\0\amp0 \end{smallmatrix}], [\begin{smallmatrix} 0\amp0\\1\amp0 \end{smallmatrix}],[\begin{smallmatrix} 0\amp0\\0\amp1 \end{smallmatrix}]\}\) be the standard basis for \(M_2(\R).\)

(a)

Define a map \(T: M_2(\R) \to \R\) by

\begin{equation*} T\left(\ba{rr}a\amp b\\c\amp d\ea\right) = a+d. \end{equation*}

The quantity \(a+d\) (the sum of the diagonal entries) is called the trace of the matrix. You may assume that \(T\) is a linear map. Find a basis for its kernel, \(K.\)

Solution.

It is easy to see that \(T\) is a surjective map, so by the rank-nullity theorem, \(\dim K = 3.\) Extracting from the standard basis, we see that \(E_{12}, E_{21} \in K\) so are part of a basis for \(K.\) We just need to add one more matrix which is not in the span of the two chosen basis vectors.

Certainly, the matrix must have the form \(\ba{rr}a\amp b\\c\amp -a\ea\text{,}\) and we need \(a\ne 0\text{,}\) otherwise our matrix is in the span of the other two vectors. But once we realize that, we may as well assume that \(b=c=0,\) so that \(\ba{rr}1\amp 0\\0\amp -1\ea\) is a nice choice, and since it is not in the span of the other two, adding it still gives us an independent set.

(b)

Now let’s consider the subspace \(S\) consisting of all symmetric matrices, those for which \(A^T = A.\) It should be clear this is a proper subspace, but what is its dimension. Actually finding a basis helps answer that question.

Hint.

If you don’t like the “brute force” force of the tack of the solution, you could take the high road and consider the space of skew-symmetric matrices, those for which \(A^T = -A\text{.}\) It is pretty easy to determine its dimension and then you can use the fact that every matrix can be written as the sum of symmetric and skew-symmetric matrix to tell you the dimension of \(S.\)

\begin{equation*} A = \frac12(A + A^T) + \frac12 (A-A^T). \end{equation*}

Solution.

Once again, it is clear that some elements of the standard basis are in \(S,\) like \(E_{11}, E_{22}\text{.}\) Since it is a proper subspace, its dimension is either 2 or 3, and a few moments thought convinces you that

\begin{equation*} \ba{rr}0\amp 1\\1\amp 0\ea = E_{12}+E_{21} \end{equation*}

is symmetric, not in the span of the other two, so forms an independent set in \(S.\) So \(\dim S = 3,\) this must be a basis for \(S.\)

(c)

Now \(K\cap S\) is also a subspace of \(M_2(\R)\text{.}\) Can we find its dimension.

Solution.

Once again, it is useful to know the dimension of the space. Certainly it is at most 3, but then not every symmetric matrix has zero trace, so it is at most two. Staring at the bases for each of \(S\) and \(K\) separately, we see that both

\begin{equation*} \ba{rr}0\amp 1\\1\amp 0\ea \text { and } \ba{rr}1\amp 0\\0\amp -1\ea \end{equation*}

are in the intersection and are clearly linearly independent, so they must be a basis.

(d)

Extend the basis you found for \(K\cap S\) to bases for \(S\) and for \(K.\)

Solution.

Since \(\dim (K\cap S) = 2\text{,}\) we need only find one matrix not in their span to give a basis for either \(K\) or \(S.\) For \(K\text{,}\) we could choose \(E_{12},\) and for \(S\) we could choose \(E_{11}.\) Knowing the dimension is clearly a powerful tool since it tells you when you are done.