Science:Math Exam Resources/Courses/MATH307/April 2012/Question 04 (b)/Solution 1

From part (a) of the question, we determined that $N(D)={\vec {1}}$ so the dimension of $N(D)$ is equal to 1, $dim(N(D))=1$.

To find $N(D^{T})$, we can use the rank-nullity theorem which states: $dim(N(A))=m-r(A)$ Where m is the number of columns in the matrix.

Remember that $r(A)=dim(R(A))$ so: $dim(N(A))=m-dim(R(A))$

So, if we solve for $dim(R(A))$, we get: $dim(R(A))=m-dim(N(A))$ $dim(R(A))=3-1$ $dim(R(A))=r(A)=2$

We can now use $dim(R(A))=2$ to solve for $dim(N(D^{T}))$ using the rank-nullity again. Finding $dim(N(D^{T}))$ does not directly give a solution to the question, but it does tell us how many vectors there are in $N(D^{T})$. The rank-nullity theorem states that: $dim(N(D^{T}))=n-r(D)$ where n is the number of rows. Plugging in for what we solved above $r(A)=2$, we get: $dim(N(D^{T}))=3-2$ $dim(N(D^{T}))=1$

This shows that there is only 1 vector in $N(D^{T})$

Now solving for what the question asks, lets look at $D^{T}$: $D^{T}={\begin{bmatrix}-1&0&-1\\1&-1&0\\0&1&1\\\end{bmatrix}}$

To find the nullspace of $D^{T}$, we have to find vectors ${\vec {v}}$ such that $D^{T}{\vec {v}}=0$. Looking at $D^{T}$, we can easily see that the only solution to this problem is: ${\vec {v}}={\begin{bmatrix}1\\1\\-1\end{bmatrix}}$

This also matches our expectation that there is only 1 vector in N(D^T).

Proof:

$rref(D)={\begin{bmatrix}1&0&1\\0&1&1\\0&0&0\\\end{bmatrix}}$

${\begin{bmatrix}x_{1}\\x_{2}\\x_{3}7\end{bmatrix}}={\begin{bmatrix}-1\\-1\\1\end{bmatrix}}={\begin{bmatrix}1\\1\\-1\end{bmatrix}}$ $N(D)={\begin{bmatrix}1\\1\\-1\end{bmatrix}}$

FINAL ANSWER ${\vec {v}}={\begin{bmatrix}1\\1\\-1\end{bmatrix}}$