Dimension

In we showed that S=\set{x-2,\,x^2-4x+4,\,x^3-6x^2+12x-8,\,x^4-8x^3+24x^2-32x+16} is a spanning set for $W=\setparts{p(x)}{p\in P_4,\ p(2)=0}$. So we can apply to $W$ with $t=4$. Here is a set of five vectors from $W$, as you may check by verifying that each is a polynomial of degree 4 or less and has $x=2$ as a root, T&=\set{p_1,\,p_2,\,p_3,\,p_4,\,p_5}\subseteq W\\ &\ \\ p_1&=x^4-2x^3+2x^2-8x+8\\ p_2&=-x^3+6x^2-5x-6\\ p_3&=2x^4-5x^3+5x^2-7x+2\\ p_4&=-x^4+4x^3-7x^2+6x\\ p_5&=4x^3-9x^2+5x-6

By we conclude that $T$ is linearly dependent, with no further computations.

It should now be plausible that Z=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{2a+b+3c+4d=0,\,-a+3b-5c-2d=0} is a subspace of the vector space $M_{22}$ (). (It is.) To find the dimension of $Z$ we must first find a basis, though any old basis will do.

First concentrate on the conditions relating $a,\,b,\,c$ and $d$. They form a homogeneous system of two equations in four variables with coefficient matrix \begin{bmatrix} 2 & 1 & 3 & 4\\ -1 & 3 & -5 & -2 \end{bmatrix}

We can row-reduce this matrix to obtain \begin{bmatrix} \leading{1} & 0 & 2 & 2\\ 0 & \leading{1} & -1 & 0 \end{bmatrix}

Rewrite the two equations represented by each row of this matrix, expressing the dependent variables ($a$ and $b$) in terms of the free variables ($c$ and $d$), and we obtain, a&=-2c-2d\\ b&=c

We can now write a typical entry of $Z$ strictly in terms of $c$ and $d$, and we can decompose the result, \begin{bmatrix}a&b\\c&d\end{bmatrix}= \begin{bmatrix}-2c-2d&c\\c&d\end{bmatrix}= \begin{bmatrix}-2c&c\\c&0\end{bmatrix}+ \begin{bmatrix}-2d&0\\0&d\end{bmatrix}= c\begin{bmatrix}-2&1\\1&0\end{bmatrix}+ d\begin{bmatrix}-2&0\\0&1\end{bmatrix}

This equation says that an arbitrary matrix in $Z$ can be written as a linear combination of the two vectors in S=\set{\begin{bmatrix}-2&1\\1&0\end{bmatrix},\,\begin{bmatrix}-2&0\\0&1\end{bmatrix}} so we know that Z=\spn{S}= \spn{\set{ \begin{bmatrix}-2&1\\1&0\end{bmatrix},\, \begin{bmatrix}-2&0\\0&1\end{bmatrix} }}

Are these two matrices (vectors) also linearly independent? Begin with a relation of linear dependence on $S$, a_1\begin{bmatrix}-2&1\\1&0\end{bmatrix}+ a_2\begin{bmatrix}-2&0\\0&1\end{bmatrix}&=\zeromatrix\\ \begin{bmatrix}-2a_1-2a_2&a_1\\a_1&a_2\end{bmatrix}&= \begin{bmatrix}0&0\\0&0\end{bmatrix}

From the equality of the two entries in the last row, we conclude that $a_1=0$, $a_2=0$. Thus the only possible relation of linear dependence is the trivial one, and therefore $S$ is linearly independent (). So $S$ is a basis for $Z$ (). Finally, we can conclude that $\dimension{Z}=2$ () since $S$ has two elements.

Let us compute the rank and nullity of A=\begin{bmatrix} 2 & -4 & -1 & 3 & 2 & 1 & -4\\ 1 & -2 & 0 & 0 & 4 & 0 & 1\\ -2 & 4 & 1 & 0 & -5 & -4 & -8\\ 1 & -2 & 1 & 1 & 6 & 1 & -3\\ 2 & -4 & -1 & 1 & 4 & -2 & -1\\ -1 & 2 & 3 & -1 & 6 & 3 & -1 \end{bmatrix}

To do this, we will first row-reduce the matrix since that will help us determine bases for the null space and column space. \begin{bmatrix} \leading{1} & -2 & 0 & 0 & 4 & 0 & 1\\ 0 & 0 & \leading{1} & 0 & 3 & 0 & -2\\ 0 & 0 & 0 & \leading{1} & -1 & 0 & -3\\ 0 & 0 & 0 & 0 & 0 & \leading{1} & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}

From this row-equivalent matrix in reduced row-echelon form we record $D=\set{1,\,3,\,4,\,6}$ and $F=\set{2,\,5,\,7}$.

For each index in $D$, creates a single basis vector. In total the basis will have $4$ vectors, so the column space of $A$ will have dimension $4$ and we write $\rank{A}=4$.

For each index in $F$, creates a single basis vector. In total the basis will have $3$ vectors, so the null space of $A$ will have dimension $3$ and we write $\nullity{A}=3$.

The matrix E=\begin{bmatrix} 0 & 4 & -1 & 2 & 2 & 3 & 1\\ 2 & -2 & 1 & -1 & 0 & -4 & -3\\ -2 & -3 & 9 & -3 & 9 & -1 & 9\\ -3 & -4 & 9 & 4 & -1 & 6 & -2\\ -3 & -4 & 6 & -2 & 5 & 9 & -4\\ 9 & -3 & 8 & -2 & -4 & 2 & 4\\ 8 & 2 & 2 & 9 & 3 & 0 & 9 \end{bmatrix} is row-equivalent to the matrix in reduced row-echelon form, \begin{bmatrix} \leading{1} & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & \leading{1} & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & \leading{1} & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \leading{1} & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & \leading{1} & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & \leading{1} & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & \leading{1} \end{bmatrix}

With $n=7$ columns and $r=7$ nonzero rows tells us the rank is $\rank{E}=7$ and the nullity is $\nullity{E}=7-7=0$.