Mathematics II · Unit 5 · 5 hrs
Vector Spaces
Exam-focused notes for Vector Spaces (Mathematics II, MTH168): what the TU syllabus asks and how it has actually been tested, with 10 solved past questions from this unit.
What this unit covers
- Vector spaces and subspaces
- Null spaces, Column spaces, and Linear transformations
- Linearly independent sets: Bases
- Coordinate systems
Null spaces, Column spaces, and Linear transformations
Define null space of a matrix A. Let then show that v belongs to the null space matrix A. $A = \begin{bmatrix} 1 & -3 & 2 \ -5 & 9 & -1 \end{bmatrix}$, $v = \begin{bmatrix} 5 \ 3 \ 2 \end{bmatrix}$ [5]
Matrix A (2×3): $$A = \begin{bmatrix} 1 & -3 & 2 \\ -5 & 9 & -1 \end{bmatrix}$$ Vector v (3×1): $$v = \begin{bmatrix} 5 \\ 3 \\ 2 \end{bmatrix}$$ All required data is present. --- For an $m \times n$ matrix $A$, the null space of $A$, denoted $N(A)$ or $\te...
Full solved answer →Define null space of a matrix $A$. Show that $v$ is in the null space of $A$, where
$$A = \begin{bmatrix} 1 & -3 & -2 \ -5 & 9 & 1 \end{bmatrix}, \qquad v = \begin{bmatrix} 5 \ 3 \ -2 \end{bmatrix}$$
The null space of an $m \times n$ matrix $A$, written $N(A)$ or $\text{Null}(A)$, is the set of all vectors $\mathbf{x}$ in $\mathbb{R}^n$ that satisfy the homogeneous equation $A\mathbf{x} = \mathbf{0}$: $$N(A) = \{\, \mathbf{x} \in \mathbb{R}^n \mid A\mat...
Full solved answer →Define null space of a matrix $A$. Let
$$A = \begin{bmatrix} 1 & -3 & -2 \ -5 & 9 & 1 \end{bmatrix}, \qquad v = \begin{bmatrix} 5 \ 3 \ -2 \end{bmatrix}$$
Then show that $v$ is in the null space of $A$.
The null space of an $m \times n$ matrix $A$, written $N(A)$ or $\text{Null}(A)$, is the set of all vectors $\mathbf{x}$ in $\mathbb{R}^n$ that satisfy the homogeneous equation $A\mathbf{x} = \mathbf{0}$: $$N(A) = \{\, \mathbf{x} \in \mathbb{R}^n \mid A\mat...
Full solved answer →Define null space. Find their basis for the null space of the matrix $A = \begin{bmatrix} 1 & 2 & 3 \ 2 & 3 & 4 \end{bmatrix}$ [5]
- Matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{bmatrix}$ (size $2 \times 3$) For an $m \times n$ matrix $A$, the null space (denoted $N(A)$ or $\text{Null}(A)$) is the set of all vectors $\mathbf{x} \in \mathbb{R}^n$ satisfying $A\mathbf{x} = \ma...
Full solved answer →Find the dimension of the null space and column space of $A = \begin{bmatrix} -3 & 6 & -1 & 1 & -7 \ 1 & -2 & 2 & 3 & -1 \ 2 & -4 & 5 & 8 & -4 \end{bmatrix}$ [5]
$$A = \begin{bmatrix} -3 & 6 & -1 & 1 & -7 \\ 1 & -2 & 2 & 3 & -1 \\ 2 & -4 & 5 & 8 & -4 \end{bmatrix}, \quad 3 \times 5 \text{ matrix}$$ Number of columns $n = 5$. Swap $R1 \leftrightarrow R2$: $$\begin{bmatrix} 1 & -2 & 2 & 3 & -1 \\ -3 & 6 & -1 & 1 & -7 ...
Full solved answer →Find the basis and dimension of Null A where $A = \begin{bmatrix} 1 & 2 & 3 & 4 \ 2 & 4 & 7 & 8 \end{bmatrix}$ [5]
$$A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 7 & 8 \end{bmatrix}$$ Matrix has 2 rows, 4 columns ($n = 4$ variables). We solve $A\mathbf{x} = \mathbf{0}$. $$\begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 7 & 8 \end{bmatrix}$$ $R2 \leftarrow R2 - 2R1$: $$\begin{b...
Full solved answer →Coordinate systems
Find the vector $\mathbf{x}$ determined by the coordinate vector $[\mathbf{x}]_\beta = \begin{bmatrix} -4 \ 8 \ 7 \end{bmatrix}$ where $\beta = \left{ \begin{bmatrix} -1 \ 2 \ 0 \end{bmatrix}, \begin{bmatrix} 3 \ -5 \ 2 \end{bmatrix}, \begin{bmatrix} 4 \ -7 \ 3 \end{bmatrix} \right}$. [5]
Basis: $$\mathbf{b}1 = \begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix}, \quad \mathbf{b}2 = \begin{bmatrix} 3 \\ -5 \\ 2 \end{bmatrix}, \quad \mathbf{b}3 = \begin{bmatrix} 4 \\ -7 \\ 3 \end{bmatrix}$$ Coordinate vector: $$[\mathbf{x}]\beta = \begin{bmatrix} -4 \...
Full solved answer →Linearly independent sets
Define linearly independent set of vectors with an example. Show that the vectors (1,4,3), (0,3,1) and (3,-5,4) are linearly independent. Do they form a basis? Justify.[10]
A set of vectors $\{\mathbf{v1}, \mathbf{v2}, \ldots, \mathbf{vn}\}$ in a vector space $V$ is linearly independent if the only solution to $$x1\mathbf{v1} + x2\mathbf{v2} + \cdots + xn\mathbf{vn} = \mathbf{0}$$ is the trivial solution $x1 = x2 = \cdots = xn...
Full solved answer →Vector spaces and subspaces
Define subspace of a vector space. Let $H = \left{\begin{bmatrix} s \ t \ 0 \end{bmatrix} : s, t \in \mathbb{R}\right}$, show that $H$ is a subspace of $\mathbb{R}^3$. [5]
A non-empty subset H of a vector space V over a field k is said to be a subspace of V if it satisfies the following conditions: 1. Zero vector: The zero vector 0 belongs to H. 2. Closure under addition: For all u, v in H, u + v is in H. 3. Closure under sca...
Full solved answer →Show that $H = {(a-3b, b-a, a, b) : a, b \in \mathbb{R}}$ is a subspace of $\mathbb{R}^4$. [5]
Given: $$H = \{(a-3b,\ b-a,\ a,\ b) : a, b \in \mathbb{R}\}$$ We rewrite a general element of H by separating the parameters $a$ and $b$: $$\begin{pmatrix} a-3b \\ b-a \\ a \\ b \end{pmatrix} = a\begin{pmatrix} 1 \\ -1 \\ 1 \\ 0 \end{pmatrix} + b\begin{pmat...
Full solved answer →Make Unit 5 stick
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