Challenge Problems for Chapter 5

Suppose \(V\) has dimension \(p\) and \(W\) has dimension \(q\) and they are each contained in a subspace, \(U\) which has dimension equal to \(n\) where \(n>\max \left ( p,q\right ).\) What are the possibilities for the dimension of \(V\cap W\)?
Remember that a linearly independent set can be extended to form a basis.

Click the arrow to see the answer.

Let \(\left \{ x_{1},\cdots ,x_{k}\right \} \) be a basis for \(V\cap W.\) Then there is a basis for \(V\) and \(W\) which are respectively
\[ \left \{ x_{1},\cdots ,x_{k},y_{k+1},\cdots ,y_{p}\right \} ,\ \left \{ x_{1},\cdots ,x_{k},z_{k+1},\cdots ,z_{q}\right \} \]
It follows that you must have \(k+p-k+q-k\leq n\) and so you must have
\[ p+q-n\leq k \]
Let \(\{\vec {v}_1, \vec {v}_2, \dots , \vec {v}_n\}\) be a basis of \(\RR ^n\). Let \(A\) be an \(n\times n\) matrix.
(a)
If \(A\) is invertible, show that \(\{A\vec {v}_1, A\vec {v}_2, \dots , A\vec {v}_n\}\) is a basis of \(\RR ^n\).
(b)
If \(\{A\vec {v}_1, A\vec {v}_2, \dots , A\vec {v}_n\}\) is a basis of \(\RR ^n\), show that \(A\) is invertible.
Show that \(\text {null}(A)=\text {null}(A^TA)\) for any real matrix \(A\).
Let \(A\) be an \(m\times n\) matrix of rank \(r\). Show that \(\text {dim}(\text {null}(A))=n-r\).
Choose a basis \(\{\vec {x}_1, \vec {x}_2, \dots , \vec {x}_k\}\) of \(\text {null}A\) and extend it to a basis \(\{\vec {x}_1, \vec {x}_2, \dots , \vec {x}_k, \vec {z}_1, \dots , \vec {z}_m\}\) of \(\RR ^n\). Show that \(\{A\vec {z}_1, \dots , A\vec {z}_m\}\) is a basis of \(\text {col}(A)\).
Let \(U\) and \(W\) be subspaces of \(\RR ^3\). If \(\dim (U)=\dim (W)=2\), and \(U\neq W\), show that \(\dim (U\cap W)=1\).
See Problem ??.

Bibliography

These problems came from Chapter 6 of Keith Nicholson’s Linear Algebra with Applications. (CC-BY-NC-SA)

W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, pp. 358–359, and p. 370.