how to find a basis of a subspace

Trouble 708

Let $A=\begin{bmatrix}
2 & iv & vi & viii \\
1 &iii & 0 & v \\
i & one & 6 & three
\end{bmatrix}$.

(a) Discover a footing for the nullspace of $A$.

(b) Find a footing for the row space of $A$.

(c) Find a basis for the range of $A$ that consists of column vectors of $A$.

(d) For each column vector which is not a footing vector that you obtained in part (c), limited it as a linear combination of the basis vectors for the range of $A$.

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Contents

• Problem 708
• Solution.
  • (a) Discover a basis for the nullspace of $A$.
  • (b) Find a footing for the row space of $A$.
  • (c) Find a ground for the range of $A$ that consists of column vectors of $A$.
  • (d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.

Solution.

Nosotros start obtain the reduced row echelon form matrix corresponding to the matrix $A$.
Nosotros reduce the matrix $A$ as follows:
\begin{marshal*}
A=\brainstorm{bmatrix}
two & 4 & 6 & 8 \\
1 &3 & 0 & 5 \\
1 & 1 & half-dozen & 3
\end{bmatrix}
\xrightarrow{\frac{1}{two}R_1}
\begin{bmatrix}
one & 2 & 3 & iv \\
1 &three & 0 & 5 \\
one & 1 & 6 & 3
\finish{bmatrix}\\[6pt] \xrightarrow[R_3-R_1]{R_2-R_1}
\brainstorm{bmatrix}
ane & 2 & 3 & 4 \\
0 &1 & -three & i \\
0 & -1 & 3 & -1
\end{bmatrix}
\xrightarrow[R_3+R_2]{R_1-2R_2}
\begin{bmatrix}
i & 0 & ix & two \\
0 &1 & -3 & 1 \\
0 & 0 & 0 & 0
\finish{bmatrix}.
\end{marshal*}
The terminal matrix is in reduced row echelon form. That is,
\[\rref(A)=\begin{bmatrix}
ane & 0 & 9 & 2 \\
0 &i & -3 & 1 \\
0 & 0 & 0 & 0
\cease{bmatrix}. \tag{*}\]

(a) Notice a basis for the nullspace of $A$.

By the computation to a higher place, nosotros run into that the full general solution of $A\mathbf{x}=\mathbf{0}$ is
\begin{align*}
x_1&=-9x_3-2x_4\\
x_2&=3x_3-x_4,
\end{align*}
where $x_3$ and $x_4$ are free variables.
Thus, the vector form solution to $A\mathbf{x}=\mathbf{0}$ is
\begin{marshal*}
\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}
=\brainstorm{bmatrix}
-9x_3-2x_4 \\
3x_3-x_4 \\
x_3 \\
x_4
\cease{bmatrix}
=x_3\begin{bmatrix}
-9 \\
three \\
ane \\
0
\end{bmatrix}+x_4\begin{bmatrix}
-2 \\
-1 \\
0 \\
ane
\end{bmatrix}.
\end{marshal*}
It follows that the nullspace of the matrix $A$ is given by
\begin{align*}
\calN(A)&=\left\{ \mathbf{x}\in \R^4 \quad \middle | \quad \mathbf{ten}= x_3\begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix}, \text{ for all } x_3, x_4 \in \R^four \right \}\\[6pt] &= \Span \left\{ \brainstorm{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix} \right \}.
\end{align*}
Thus, the set up
\[\left\{ \begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix} \right \}\] is a spanning gear up for the nullspace $\calN(A)$.
It is straightforward to see that this fix is linearly independent, and hence information technology is a basis for $\calN(A)$.

(b) Notice a basis for the row infinite of $A$.

Recall that the nonzero rows of $\rref(A)$ class a footing for the row space of $A$.
Thus,
\[\left\{\begin{bmatrix}
i \\
0 \\
9 \\
2
\end{bmatrix}, \quad \begin{bmatrix}
0 \\
one \\
-three \\
1
\end{bmatrix} \right \}\] is a basis for the row infinite of $A$.

(c) Find a basis for the range of $A$ that consists of column vectors of $A$.

Recall that past the leading 1 method, the columns of $A$ corresponding to columns of $\rref(A)$ that contain leading 1 entries grade a basis for the range $\calR(A)$ of $A$.
From (*), nosotros see that the start and the second columns incorporate the leading ane entries. Thus,
\[\left\{\begin{bmatrix}
2 \\
1 \\
1
\end{bmatrix}, \begin{bmatrix}
four \\
3 \\
1
\cease{bmatrix}\right \}\] is a basis for the range $\calR(A)$ of $A$.

(d) For each column vector which is not a basis vector that you obtained in part (c), limited it as a linear combination of the basis vectors for the range of $A$.

Permit the states write $A_1, A_2, A_3$, and $A_4$ for the cavalcade vectors of the matrix $A$.
In part (c), nosotros showed that $\{A_1, A_2\}$ is a footing for the range $\calR(A)$.
Thus, we demand to express the vectors $A_3$ and $A_4$ as a linear combination of $A_1$ and $A_2$, respectively.

A shortcut is to annotation that the entries of tertiary cavalcade vector of $\rref(A)$ give the coefficients of the linear combination for $A_3$. That is, we take
\[A_3=9A_1-3A_2.\] Similarly, the entries of the fourth column of $\rref(A)$ yield
\[A_4=2A_1+A_2.\]

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Source: https://yutsumura.com/how-to-find-a-basis-for-the-nullspace-row-space-and-range-of-a-matrix/

Posted by: reedyhadis1955.blogspot.com

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