Answers to Column Space Practise Problems for Discover Linear Algebra

1. 1. The RREF of this matrix is the identity, so the two columns $c_{1} = [\begin{matrix} 1 \\ 4 \end{matrix}], c_{2} = [\begin{array}{r} 3 \\ - 6 \end{array}]$ are a basis for the column space.
  2. There are no other columns — all the columns ended being part of the basis.
  3. The vector equation $k_{1} c_{1} + k_{2} c_{2} = b$ leads to a linear system with $A$ as the coefficient matrix and $b$ as the "equals" column. So augment $A$ with $b$ and row reduce: $[\begin{array}{rrr} 1 & 3 & - 2 \\ 4 & - 6 & 10 \end{array}] \to [\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & - 1 \end{array}] .$ So $b = c_{1} - c_{2}$ .
  4. Since $b$ is a linear combination of the columns of $A$ , it is in the column space of $A$ , and so the system $A x = b$ is consistent.
2. 1. The RREF of this matrix is $[\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix}] .$ Since only the first two columns have leading ones, only the first two columns $\begin{aligned} c_{1} & = [\begin{array}{c} 1 \\ 1 \\ 2 \end{array}], & c_{2} = [\begin{array}{c} 1 \\ 0 \\ 1 \end{array}] \end{aligned}$ are required for a basis for the column space.
  2. In the RREF above, the third column is equal to the sum of the first two columns. So the same relationship will hold for the columns of $A$ : $c_{3} = c_{1} + c_{2}$ .
  3. The vector equation $k_{1} c_{1} + k_{2} c_{2} = b$ leads to linear system with augmented matrix $[\begin{array}{rrr} 1 & 1 & - 1 \\ 1 & 0 & 0 \\ 2 & 1 & 2 \end{array}] .$ Notice that this is just the first two columns of $A$ augmented with $b$ . Row reduce: $[\begin{array}{rrr} 1 & 1 & - 1 \\ 1 & 0 & 0 \\ 2 & 1 & 2 \end{array}] \to [\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] .$ From the last row we can see that there is no solution. Or, alternatively, we can interpret the leading one in the last column as indicating that $b$ is independent from the two basis vectors for the column space of $A$ . So it is not possible to express $b$ as a linear combination of the basis vectors from (i).
  4. Since $b$ is not in the column space of $A$ , the system $A x = b$ is inconsistent.
3. 1. The RREF of this matrix is the identity, so all three columns are independent, and all three are required to form a basis for the column space.
  2. There are no other columns — we used all three columns for the basis in part i.
  3. The vector equation $k_{1} c_{1} + k_{2} c_{2} + k_{3} c_{3} = b$ leads to a linear system with $A$ as the coefficient matrix and $b$ as the "equals" column. So augment $A$ with $b$ and row reduce: $[\begin{array}{rrrr} 1 & - 1 & 1 & 5 \\ 9 & 3 & 1 & 1 \\ 1 & 1 & 1 & - 1 \end{array}] \to [\begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & - 3 \\ 0 & 0 & 1 & 1 \end{array}] .$ So $b = c_{1} - 3 c_{2} + c_{3}$ .
  4. Since $b$ is in the column space of $A$ , the system $A x = b$ is consistent.
4. 1. We don't need to row reduce this one — we can see that the third column is the negative of the second, so it is dependent and should be discarded. That leaves us with the first two columns, which are independent because neither is a scalar multiple of the other. So the first two columns of $A$ form a basis for the column space.
  2. We already determined that $c_{3} = 0 c_{1} - c_{2}$ .
  3. Again, we don't need to row reduce this one — we can easily see that $b$ is equal to the sum of the first and third columns of $A$ . Since we determined in part i that the third column was the negative of the second, we then have $b = c_{1} - c_{2}$ .
  4. Since $b$ is in the column space of $A$ , the system $A x = b$ is consistent.
5. 1. The RREF for this matrix is the identity, so all four columns are required to form a basis for the column space of $A$ .
  2. There are no other columns — all columns were required in our basis in part i.
  3. The vector equation $k_{1} c_{1} + k_{2} c_{2} + k_{3} c_{3} + k_{4} c_{4} = b$ leads to a linear system with $A$ as the coefficient matrix and $b$ as the "equals" column. So augment $A$ with $b$ and row reduce: $[\begin{array}{rrrrr} 1 & 2 & 0 & 1 & 4 \\ 0 & 1 & 2 & 1 & 3 \\ 1 & 2 & 1 & 3 & 5 \\ 0 & 1 & 2 & 2 & 7 \end{array}] \to [\begin{array}{rrrrr} 1 & 0 & 0 & 0 & - 26 \\ 0 & 1 & 0 & 0 & 13 \\ 0 & 0 & 1 & 0 & - 7 \\ 0 & 0 & 0 & 1 & 4 \end{array}] .$ So $b = - 26 c_{1} + 13 c_{2} - 7 c_{3} + 4 c_{4} .$
  4. Since $b$ is in the column space of $A$ , the system $A x = b$ is consistent.
We can use column space to answer each of these.
1. Place the four vectors in order as columns in a $4 \times 4$ matrix and reduce. The RREF has leading ones in columns 1 and 2, so a basis for this subspace is ${v_{1}, v_{2}}$ .
2. Place the four vectors in order as columns in a $4 \times 4$ matrix and reduce. The RREF has leading ones in columns 1 and 3, so a basis for this subspace is ${v_{1}, v_{3}}$ .
3. Place the five vectors in order as columns in a $4 \times 5$ matrix and reduce. The RREF has leading ones in columns 1, 2, and 4, so a basis for this subspace is ${v_{1}, v_{2}, v_{4}}$ .

:: Answers to column space practise problems