Let \(\vec {v}=\begin {bmatrix} 5\\ 3\\ -6 \end {bmatrix} ,\vec {v}_{1}=\begin {bmatrix} -1\\ 1\\ 2 \end {bmatrix} ,\vec {v}_{2}=\begin {bmatrix} 3\\ 1\\ -4 \end {bmatrix} \) be in \(\mathbb {R} ^{3}\). Let \(W=span\left (v_{1},v_{2}\right ) \). Determine if \(\vec {v}\) is in \(W\)
Solution
\(\vec {v}\) is in \(W\) if \(\vec {v}\) can be expressed as a linear combination of basis \(\vec {v}_{1},\vec {v}_{2}\). To find if this is the case, we solve\[ c_{1}\vec {v}_{1}+c_{2}\vec {v}_{2}=\vec {v}\] If a solution exists, then \(\vec {v}\) is in \(W\). Writing the above in matrix form gives\begin {align} c_{1}\begin {bmatrix} -1\\ 1\\ 2 \end {bmatrix} +c_{2}\begin {bmatrix} 3\\ 1\\ -4 \end {bmatrix} & =\begin {bmatrix} 5\\ 3\\ -6 \end {bmatrix} \nonumber \\\begin {bmatrix} -1 & 3\\ 1 & 1\\ 2 & -4 \end {bmatrix}\begin {bmatrix} c_{1}\\ c_{2}\end {bmatrix} & =\begin {bmatrix} 5\\ 3\\ -6 \end {bmatrix} \tag {1} \end {align}
Hence the augmented matrix is\[\begin {bmatrix} -1 & 3 & 5\\ 1 & 1 & 3\\ 2 & -4 & -6 \end {bmatrix} \] \(R_{2}=R_{2}+R_{1}\) gives\[\begin {bmatrix} -1 & 3 & 5\\ 0 & 4 & 8\\ 2 & -4 & -6 \end {bmatrix} \] \(R_{3}=R_{3}+2R_{1}\) gives\[\begin {bmatrix} -1 & 3 & 5\\ 0 & 4 & 8\\ 0 & 2 & 4 \end {bmatrix} \] \(R_{3}=R_{3}-\frac {R_{2}}{2}\) gives\[\begin {bmatrix} -1 & 3 & 5\\ 0 & 4 & 8\\ 0 & 0 & 0 \end {bmatrix} \] The pivots columns are the first two columns. The system (1) now becomes\[\begin {bmatrix} -1 & 3\\ 0 & 4\\ 0 & 0 \end {bmatrix}\begin {bmatrix} c_{1}\\ c_{2}\end {bmatrix} =\begin {bmatrix} 5\\ 8\\ 0 \end {bmatrix} \] Last row provides no information as It just says \(0=0\). Second row gives \(4c_{2}=8\) or \(c_{2}=2\). First row gives \(-c_{1}+3c_{2}=5\) or \(-c_{1}=5-3\relax (2) \) or \(c_{1}=1\). Hence the solution is \[\begin {bmatrix} c_{1}\\ c_{2}\end {bmatrix} =\begin {bmatrix} 1\\ 2 \end {bmatrix} \] Since a solution is found, this means \(\vec {v}\) can be expressed as a linear combination of \(\vec {v}_{1},\vec {v}_{2}\) given by
\[ \vec {v}_{1}+2\vec {v}_{2}=\vec {v}\] Hence \(\vec {v}\) is in \(W\).
Determine the component vector of \(p\relax (x) =-4x^{2}+2x+6\) in the given vector space \(V=P_{2}\left ( \mathbb {R} \right ) \) relative to the given ordered basis \(B=\left \{ x^{2}+x,2+2x,1\right \} \)
Solution
The problem is asking us to find \(c_{1},c_{2},c_{3}\) such that \begin {align*} c_{1}\left (x^{2}+x\right ) +c_{2}\left (2+2x\right ) +c_{3}\left ( 1\right ) & =p\relax (x) \\ & =-4x^{2}+2x+6 \end {align*}
Expanding gives\begin {align*} c_{1}x^{2}+c_{1}x+2c_{2}+2c_{2}x+c_{3} & =-4x^{2}+2x+6\\ x^{2}\left (c_{1}\right ) +x\left (c_{1}+2c_{2}\right ) +\left ( 2c_{2}+c_{3}\right ) & =-4x^{2}+2x+6 \end {align*}
For these to be equal, the corresponding coefficients of the polynomials must be the same. Therefore equating coefficients of each power of \(x\) gives\begin {align*} c_{1} & =-4\\ c_{1}+2c_{2} & =2\\ 2c_{2}+c_{3} & =6 \end {align*}
Or\begin {align*} c_{1} & =-4\\ -4+2c_{2} & =2\\ 2c_{2}+c_{3} & =6 \end {align*}
Or\begin {align*} c_{1} & =-4\\ c_{2} & =3\\ 2c_{2}+c_{3} & =6 \end {align*}
Or\begin {align*} c_{1} & =-4\\ c_{2} & =3\\ 2\relax (3) +c_{3} & =6 \end {align*}
Or\begin {align*} c_{1} & =-4\\ c_{2} & =3\\ c_{3} & =0 \end {align*}
Hence the component vector of \(p\relax (x) \) is \(\left \{ -4,3,0\right \} \) relative to the basis \(B\).
Determine if the given linear transformation is a) one-to-one and b) onto. Justify your answer.\[ T\left (x,y,z\right ) =\left (x,x+y+z\right ) \] Solution
In Matrix forum the above is\[ A\begin {bmatrix} x\\ y\\ z \end {bmatrix} =\begin {bmatrix} x\\ x+y+z \end {bmatrix} \] Therefore \(A\) matrix must have dimensions \(2\times 3\). The above becomes\[\begin {bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\end {bmatrix}\begin {bmatrix} x\\ y\\ z \end {bmatrix} =\begin {bmatrix} x\\ x+y+z \end {bmatrix} \] First row gives \(a_{11}x+a_{12}y+a_{13}z=x\). Equating coefficients of the polynomials on each side gives \(a_{11}=1,a_{12}=0,a_{13}=0\). Second row gives \(a_{21}x+a_{22}y+a_{23}z=x+y+z\) which implies that \(a_{21}=1,a_{22}=1,a_{23}=1\). Hence the matrix \(A\) is\[ A=\begin {bmatrix} 1 & 0 & 0\\ 1 & 1 & 1 \end {bmatrix} \] Now that we have found the matrix representation of the linear transformation \(T\), we can answer parts a and b.
Using Theorem 6.4.8 which says for the linear transformation \(T:V\rightarrow W\)
one-to-one iff \(\ker \relax (T) =\left \{ \vec {0}\right \} \)
onto iff \(Rng\relax (T) =W\)
In this problem \(V\) is \(\mathbb {R} ^{3}\) and \(W\) is \(\mathbb {R} ^{2}\). To show one-to-one, we need to find \(\ker \relax (T) \) by solving \(A\vec {x}=\vec {0}\) and check if it is the zero vector or not.\begin {equation} \begin {bmatrix} 1 & 0 & 0\\ 1 & 1 & 1 \end {bmatrix}\begin {bmatrix} x\\ y\\ z \end {bmatrix} =\begin {bmatrix} 0\\ 0 \end {bmatrix} \tag {2} \end {equation} The augmented matrix is\[\begin {bmatrix} 1 & 0 & 0 & 0\\ 1 & 1 & 1 & 0 \end {bmatrix} \] \(R_{2}=R_{2}-R_{1}\) gives\[\begin {bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 \end {bmatrix} \] Therefore the base variables are \(x,y\) and the free variable is \(z=t\). Hence (2) becomes\[\begin {bmatrix} 1 & 0 & 0\\ 0 & 1 & 1 \end {bmatrix}\begin {bmatrix} x\\ y\\ z \end {bmatrix} =\begin {bmatrix} 0\\ 0 \end {bmatrix} \] First row gives \(x=0\) and second row gives \(y+z=0\) or \(y=-t\) Therefore the solution is\begin {align*} \begin {bmatrix} x\\ y\\ z \end {bmatrix} & =\begin {bmatrix} 0\\ -t\\ t \end {bmatrix} \\ & =t\begin {bmatrix} 0\\ -1\\ 1 \end {bmatrix} \end {align*}
Therefore null-space is \(\left \{ \begin {bmatrix} 0\\ -1\\ 1 \end {bmatrix} \right \} \). Since null-space is not the zero vector, then \(T\) is not one-to-one. To check if it is onto, the \(Rng\relax (T) \) is the column space of \(A\). From the above RREF, we found that the first two columns are the pivot columns. These correspond to the first two columns of \(A\). Therefore\[ Rng\relax (T) =\left \{ \vec {v}\in \mathbb {R} ^{3}:\vec {v}=c_{1}\begin {bmatrix} 1\\ 1 \end {bmatrix} +c_{2}\begin {bmatrix} 0\\ 1 \end {bmatrix} ,c_{1},c_{2}\in \mathbb {R} \right \} \] The question now is, does the above span all of \(W\) which is \(\mathbb {R} ^{2}\)? it is clear it does, since \(\begin {bmatrix} 1\\ 1 \end {bmatrix} ,\begin {bmatrix} 0\\ 1 \end {bmatrix} \) are linearly independent of each others and any two linearly independent vectors in \(\mathbb {R} ^{2}\) span all of \(\mathbb {R} ^{2}\). Hence it is onto.
If we need to show this, then this can be done by solving\begin {align*} c_{1}\begin {bmatrix} 1\\ 1 \end {bmatrix} +c_{2}\begin {bmatrix} 0\\ 1 \end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \\\begin {bmatrix} 1 & 0\\ 1 & 1 \end {bmatrix}\begin {bmatrix} c_{1}\\ c_{2}\end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \end {align*}
First row gives \(c_{1}=0\) and second row gives \(c_{2}=0\). Hence only solution to \(c_{1}\begin {bmatrix} 1\\ 1 \end {bmatrix} +c_{2}\begin {bmatrix} 0\\ 1 \end {bmatrix} =\begin {bmatrix} 0\\ 0 \end {bmatrix} \) is \(c_{1}=c_{2}=0\). Therefore these two vectors are linearly independent vectors in \(\mathbb {R} ^{2}\). Hence \(Rng\relax (T) \) is all \(W\). Therefore \(T\) is onto.
Given the linear transformation \(T:\mathbb {R} ^{3}\rightarrow \mathbb {R} ^{3}\) defined by the equation, find the standard matrix for the inverse transformation \(T^{-1}\)\begin {align*} w_{1} & =x_{1}+4x_{2}-x_{3}\\ w_{2} & =2x_{1}+7x_{2}+x_{3}\\ w_{3} & =x_{1}+3x_{2} \end {align*}
Solution
In matrix form the above is\[\begin {bmatrix} 1 & 4 & -1\\ 2 & 7 & 1\\ 1 & 3 & 0 \end {bmatrix}\begin {bmatrix} x_{1}\\ x_{2}\\ x_{3}\end {bmatrix} =\begin {bmatrix} w_{1}\\ w_{2}\\ w_{2}\end {bmatrix} \] Where \(A=\begin {bmatrix} 1 & 4 & -1\\ 2 & 7 & 1\\ 1 & 3 & 0 \end {bmatrix} \) represents the linear transformation \(T\). Hence \(T^{-1}\) is represented by \(A^{-1}\). Therefore we need to find \(A^{-1}\). Setting up the augmented matrix for finding the inverse is setup by adding the identity matrix to the right half as follows\[\begin {bmatrix} 1 & 4 & -1 & 1 & 0 & 0\\ 2 & 7 & 1 & 0 & 1 & 0\\ 1 & 3 & 0 & 0 & 0 & 1 \end {bmatrix} \] \(R_{2}=R_{2}-2R_{1}\)\[\begin {bmatrix} 1 & 4 & -1 & 1 & 0 & 0\\ 0 & -1 & 3 & -2 & 1 & 0\\ 1 & 3 & 0 & 0 & 0 & 1 \end {bmatrix} \] \(R_{3}=R_{3}-R_{1}\)\[\begin {bmatrix} 1 & 4 & -1 & 1 & 0 & 0\\ 0 & -1 & 3 & -2 & 1 & 0\\ 0 & -1 & 1 & -1 & 0 & 1 \end {bmatrix} \] \(R_{3}=R_{3}-R_{2}\)\[\begin {bmatrix} 1 & 4 & -1 & 1 & 0 & 0\\ 0 & -1 & 3 & -2 & 1 & 0\\ 0 & 0 & -2 & 1 & -1 & 1 \end {bmatrix} \] \(R_{3}=\frac {-R_{3}}{2},R_{2}=-R_{2}\)\[\begin {bmatrix} 1 & 4 & -1 & 1 & 0 & 0\\ 0 & 1 & -3 & 2 & -1 & 0\\ 0 & 0 & 1 & \frac {-1}{2} & \frac {1}{2} & \frac {-1}{2}\end {bmatrix} \] Now we start the reduced Echelon phase. \(R_{2}=R_{2}+3R_{3}\)\[\begin {bmatrix} 1 & 4 & -1 & 1 & 0 & 0\\ 0 & 1 & 0 & 2+3\left (\frac {-1}{2}\right ) & -1+3\left (\frac {1}{2}\right ) & -\frac {3}{2}\\ 0 & 0 & 1 & \frac {-1}{2} & \frac {1}{2} & \frac {-1}{2}\end {bmatrix} =\begin {bmatrix} 1 & 4 & -1 & 1 & 0 & 0\\ 0 & 1 & 0 & \frac {1}{2} & \frac {1}{2} & -\frac {3}{2}\\ 0 & 0 & 1 & -\frac {1}{2} & \frac {1}{2} & -\frac {1}{2}\end {bmatrix} \] \(R_{1}=R_{1}+R_{3}\)\[\begin {bmatrix} 1 & 4 & 0 & \frac {1}{2} & \frac {1}{2} & -\frac {1}{2}\\ 0 & 1 & 0 & \frac {1}{2} & \frac {1}{2} & -\frac {3}{2}\\ 0 & 0 & 1 & -\frac {1}{2} & \frac {1}{2} & -\frac {1}{2}\end {bmatrix} \] \(R_{1}=R_{1}-4R_{2}\)\[\begin {bmatrix} 1 & 0 & 0 & \frac {1}{2}-4\left (\frac {1}{2}\right ) & \frac {1}{2}-4\left ( \frac {1}{2}\right ) & -\frac {1}{2}-4\left (-\frac {3}{2}\right ) \\ 0 & 1 & 0 & \frac {1}{2} & \frac {1}{2} & -\frac {3}{2}\\ 0 & 0 & 1 & -\frac {1}{2} & \frac {1}{2} & -\frac {1}{2}\end {bmatrix} =\begin {bmatrix} 1 & 0 & 0 & -\frac {3}{2} & -\frac {3}{2} & \frac {11}{2}\\ 0 & 1 & 0 & \frac {1}{2} & \frac {1}{2} & -\frac {3}{2}\\ 0 & 0 & 1 & -\frac {1}{2} & \frac {1}{2} & -\frac {1}{2}\end {bmatrix} \] Now that the left half above is the identity matrix, then the right half is \(A^{-1}\). Therefore\[ A^{-1}=\begin {bmatrix} -\frac {3}{2} & -\frac {3}{2} & \frac {11}{2}\\ \frac {1}{2} & \frac {1}{2} & -\frac {3}{2}\\ -\frac {1}{2} & \frac {1}{2} & -\frac {1}{2}\end {bmatrix} \] Which is the standard matrix for the representation of \(T^{-1}\)
Answer the following questions by writing TRUE or FALSE.
Solution
a
The rank of a matrix equals the dimension of its column space. TRUE
b
The number of variables in the equations \(Ax=0\) equals the dimension of the nullsapce of \(A\). FALSE
c
If \(A\) is \(3\times 5\) matrix and \(T\) is a transformation defined by \(T\relax (x) =Ax\), then the domain of \(T\) is \(\mathbb {R} ^{3}\). FALSE
d
If a \(4\times 7\) matrix \(A\) has four pivot columns then the \(nullity\relax (A) =3\). TRUE
e
A linearly independent set in a subspace \(H\) is a basis for \(H\). FALSE