Linear Transformation: Definition, Example, and Solution

Definition

A transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is a linear transformation if for all vectors $\mathbf{u}, \mathbf{v} \in \mathbb{R}^n$ and all scalars $c$:

1. $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$
2. $T(c\mathbf{u}) = c,T(\mathbf{u})$

Example: For any $m \times n$ matrix $A$, the map $T(\mathbf{x}) = A\mathbf{x}$ is a linear transformation.

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Step 1: Given Data

$$A = \begin{bmatrix} 1 & -3 \ 3 & 5 \ -1 & 7 \end{bmatrix}, \quad v = \begin{bmatrix} 2 \ -1 \end{bmatrix}, \quad b = \begin{bmatrix} 3 \ 2 \ 4 \end{bmatrix}, \quad x = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}$$

$T(\mathbf{x}) = A\mathbf{x}$.

Note: Since $A$ is $3 \times 2$, the map is actually $T:\mathbb{R}^2 \to \mathbb{R}^3$ (the problem statement's "$\mathbb{R}^2$" codomain is a typo). I use $\mathbb{R}^3$ as the codomain.

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Step 2: Solve

Part (a): $T(v) = Av$

$$T(v) = \begin{bmatrix} 1 & -3 \ 3 & 5 \ -1 & 7 \end{bmatrix}\begin{bmatrix} 2 \ -1 \end{bmatrix}$$

• Row 1: $(1)(2) + (-3)(-1) = 2 + 3 = 5$
• Row 2: $(3)(2) + (5)(-1) = 6 - 5 = 1$
• Row 3: $(-1)(2) + (7)(-1) = -2 - 7 = -9$

$$\boxed{T(v) = \begin{bmatrix} 5 \ 1 \ -9 \end{bmatrix}}$$

Part (b): Solve $Ax = b$

$$\left[\begin{array}{cc|c} 1 & -3 & 3 \ 3 & 5 & 2 \ -1 & 7 & 4 \end{array}\right]$$

$R_2 \to R_2 - 3R_1$, $R_3 \to R_3 + R_1$:

$$\left[\begin{array}{cc|c} 1 & -3 & 3 \ 0 & 14 & -7 \ 0 & 4 & 7 \end{array}\right]$$

$R_2 \to \tfrac{1}{14}R_2$:

$$\left[\begin{array}{cc|c} 1 & -3 & 3 \ 0 & 1 & -\tfrac{1}{2} \ 0 & 4 & 7 \end{array}\right]$$

$R_3 \to R_3 - 4R_2$:

$$R_3 = \left[0,; 4 - 4(1),; 7 - 4\left(-\tfrac{1}{2}\right)\right] = \left[0,; 0,; 7 + 2\right] = [0,;0,;9]$$

$$\left[\begin{array}{cc|c} 1 & -3 & 3 \ 0 & 1 & -\tfrac{1}{2} \ 0 & 0 & 9 \end{array}\right]$$

The last row states $0 = 9$, which is impossible.

Conclusion: The system $Ax = b$ is inconsistent. There is no $x \in \mathbb{R}^2$ whose image under $T$ equals $b = [3,;2,;4]^T$. Equivalently, $b$ is not in the range (column space) of $A$.

Verification. A candidate sometimes quoted for this system is $x_1 = 1.5,\ x_2 = -0.5$. Testing it: $$A\begin{bmatrix}1.5\-0.5\end{bmatrix} = \begin{bmatrix}(1)(1.5)+(-3)(-0.5)\(3)(1.5)+(5)(-0.5)\(-1)(1.5)+(7)(-0.5)\end{bmatrix} = \begin{bmatrix}3\2\-5\end{bmatrix} \neq \begin{bmatrix}3\2\4\end{bmatrix}.$$

The first two equations are satisfied but the third gives $-5 \ne 4$, confirming there is no solution.

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Final Answers

• (a) $T(v) = [5,\ 1,\ -9]^T$
• (b) No solution exists; $Ax = b$ is inconsistent ($b$ is not in the column space of $A$).

Vector Equality and Operations

STEP 1 - EXTRACT: Given Data

$$\mathbf{u} = \begin{bmatrix} 1 \ -3 \ 2 \end{bmatrix}, \quad \mathbf{v} = \begin{bmatrix} 1 \ -1 \ 3 \end{bmatrix}$$

Required:

• Condition for equality of two column vectors in $\mathbb{R}^2$ with an example
• Compute $\mathbf{u} + 3\mathbf{v}$ and $\mathbf{u} - 2\mathbf{v}$

Note: The given vectors $\mathbf{u}$ and $\mathbf{v}$ each have 3 components, so they actually lie in $\mathbb{R}^3$, not $\mathbb{R}^2$. The theory part asks about $\mathbb{R}^2$, but the computation uses $\mathbb{R}^3$ vectors. The operations are still well defined (both same size), so we proceed as given.

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STEP 2 - SOLVE

Part 1: When are two column vectors in $\mathbb{R}^2$ equal?

Two column vectors are equal if and only if their corresponding components are equal (this requires them to have the same number of entries).

For $\mathbf{a} = \begin{bmatrix} a_1 \ a_2 \end{bmatrix}$ and $\mathbf{b} = \begin{bmatrix} b_1 \ b_2 \end{bmatrix}$:

$$\mathbf{a} = \mathbf{b} \iff a_1 = b_1 \text{ and } a_2 = b_2$$

Example:

$$\begin{bmatrix} 2 \ 7 \end{bmatrix} = \begin{bmatrix} 2 \ 7 \end{bmatrix}$$

is an equality since $2 = 2$ and $7 = 7$. However,

$$\begin{bmatrix} 2 \ 7 \end{bmatrix} \neq \begin{bmatrix} 7 \ 2 \end{bmatrix}$$

since the components (though the same set of numbers) are not in matching positions.

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Part 2: Compute $\mathbf{u} + 3\mathbf{v}$

Scalar multiply:

$$3\mathbf{v} = \begin{bmatrix} 3(1) \ 3(-1) \ 3(3) \end{bmatrix} = \begin{bmatrix} 3 \ -3 \ 9 \end{bmatrix}$$

Add:

$$\mathbf{u} + 3\mathbf{v} = \begin{bmatrix} 1+3 \ -3+(-3) \ 2+9 \end{bmatrix} = \begin{bmatrix} 4 \ -6 \ 11 \end{bmatrix}$$

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Compute $\mathbf{u} - 2\mathbf{v}$

Scalar multiply:

$$2\mathbf{v} = \begin{bmatrix} 2(1) \ 2(-1) \ 2(3) \end{bmatrix} = \begin{bmatrix} 2 \ -2 \ 6 \end{bmatrix}$$

Subtract:

$$\mathbf{u} - 2\mathbf{v} = \begin{bmatrix} 1-2 \ -3-(-2) \ 2-6 \end{bmatrix} = \begin{bmatrix} -1 \ -1 \ -4 \end{bmatrix}$$

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Final Results

$$\mathbf{u} + 3\mathbf{v} = \begin{bmatrix} 4 \ -6 \ 11 \end{bmatrix}, \qquad \mathbf{u} - 2\mathbf{v} = \begin{bmatrix} -1 \ -1 \ -4 \end{bmatrix}$$

Image of Vectors Under Linear Transformation $T(x) = Ax$

STEP 1 - Given Data

$$A = \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix}, \quad u = \begin{bmatrix} 1 \ -3 \end{bmatrix}, \quad v = \begin{bmatrix} 1 \ 5 \end{bmatrix}$$

Transformation: $T(x) = Ax$, with $T:\mathbb{R}^2 \to \mathbb{R}^2$.

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STEP 2 - Solve

Finding $T(u)$

$$T(u) = Au = \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix} \begin{bmatrix} 1 \ -3 \end{bmatrix}$$

• Row 1: $(0)(1) + (1)(-3) = -3$
• Row 2: $(-1)(1) + (0)(-3) = -1$

$$T(u) = \begin{bmatrix} -3 \ -1 \end{bmatrix}$$

Finding $T(v)$

$$T(v) = Av = \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix} \begin{bmatrix} 1 \ 5 \end{bmatrix}$$

• Row 1: $(0)(1) + (1)(5) = 5$
• Row 2: $(-1)(1) + (0)(5) = -1$

$$T(v) = \begin{bmatrix} 5 \ -1 \end{bmatrix}$$

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Summary

Note on geometry: The matrix $A = \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix}$ rotates each vector by $90°$ clockwise. (A standard $90°$ counterclockwise rotation matrix is $\begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$; the transpose form here gives clockwise rotation.)

Null Space of a Matrix

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 $A\mathbf{x} = \mathbf{0}$:

$$N(A) = {, \mathbf{x} \in \mathbb{R}^n \mid A\mathbf{x} = \mathbf{0} ,}$$

so the null space is exactly the solution set of $A\mathbf{x} = \mathbf{0}$. Its vectors carry as many components as $A$ has columns, while the zero vector on the right carries as many components as $A$ has rows.

The null space is always a subspace of $\mathbb{R}^n$. It contains the zero vector because $A\mathbf{0} = \mathbf{0}$, it is closed under addition because $A\mathbf{u} = \mathbf{0}$ and $A\mathbf{v} = \mathbf{0}$ give

$$A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v} = \mathbf{0}$$

and it is closed under scalar multiplication because $A\mathbf{u} = \mathbf{0}$ gives

$$A(c\mathbf{u}) = c(A\mathbf{u}) = \mathbf{0}$$

Its dimension is called the nullity of $A$, and by the rank-nullity theorem the rank and the nullity add up to $n$, the number of columns.

Showing that $v$ lies in $\text{Null}(A)$

We are given

$$A = \begin{bmatrix} 1 & -3 & -2 \ -5 & 9 & 1 \end{bmatrix}, \qquad v = \begin{bmatrix} 5 \ 3 \ -2 \end{bmatrix}$$

Here $A$ has three columns, so the vectors of its null space live in $\mathbb{R}^3$, and $v$ has three components, which makes the test meaningful. Membership needs no row reduction at all, because it is settled by the single defining condition $Av = \mathbf{0}$. Forming the product row by row,

$$Av = \begin{bmatrix} 1 & -3 & -2 \ -5 & 9 & 1 \end{bmatrix} \begin{bmatrix} 5 \ 3 \ -2 \end{bmatrix}$$

The first row gives

$$(1)(5) + (-3)(3) + (-2)(-2) = 5 - 9 + 4 = 0$$

and the second row gives

$$(-5)(5) + (9)(3) + (1)(-2) = -25 + 27 - 2 = 0$$

so that

$$Av = \begin{bmatrix} 0 \ 0 \end{bmatrix} = \mathbf{0}$$

Since $Av = \mathbf{0}$, the vector $v$ satisfies the defining condition of the null space. Therefore $v \in \text{Null}(A)$, which is what the question asked us to show.

Unit Vector of u = (1, −2, 2, 3)

Step 1: Given data

$$u = (1,\ -2,\ 2,\ 3)$$

Step 2: Definition

The unit vector in the direction of $u$ is:

$$\hat{u} = \frac{u}{|u|}, \qquad |u| = \sqrt{u_1^2 + u_2^2 + u_3^2 + u_4^2}$$

Step 3: Compute the norm

$$|u| = \sqrt{(1)^2 + (-2)^2 + (2)^2 + (3)^2} = \sqrt{1 + 4 + 4 + 9} = \sqrt{18} = 3\sqrt{2}$$

Step 4: Compute the unit vector

$$\hat{u} = \frac{1}{3\sqrt{2}}(1,\ -2,\ 2,\ 3) = \left(\frac{1}{3\sqrt{2}},\ \frac{-2}{3\sqrt{2}},\ \frac{2}{3\sqrt{2}},\ \frac{3}{3\sqrt{2}}\right)$$

Rationalizing:

$$\hat{u} = \left(\frac{\sqrt{2}}{6},\ \frac{-\sqrt{2}}{3},\ \frac{\sqrt{2}}{3},\ \frac{\sqrt{2}}{2}\right)$$

Step 5: Verification

$$|\hat{u}| = \sqrt{\frac{1}{18} + \frac{4}{18} + \frac{4}{18} + \frac{9}{18}} = \sqrt{\frac{18}{18}} = 1 \checkmark$$

Hence the unit vector is confirmed.

STEP 1 - Given Data

Three signals to test for linear independence:

$$f_1(k) = 1^k, \quad f_2(k) = (-2)^k, \quad f_3(k) = 3^k$$

No other numeric data required.

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STEP 2 - Solution

Definition

Signals ${f_1, f_2, f_3}$ are linearly independent if

$$C_1 f_1(k) + C_2 f_2(k) + C_3 f_3(k) = 0 \quad \text{for all } k$$

implies $C_1 = C_2 = C_3 = 0$.

Form the combination

$$C_1 (1)^k + C_2 (-2)^k + C_3 (3)^k = 0 \quad \text{for all } k$$

Evaluate at three sample values of $k$.

$k = 0$: $$C_1 + C_2 + C_3 = 0 \quad (1)$$

$k = 1$: $$C_1 - 2C_2 + 3C_3 = 0 \quad (2)$$

$k = 2$: $$C_1 + 4C_2 + 9C_3 = 0 \quad (3)$$

Casoratian (determinant) test

This is a homogeneous system $A\mathbf{C} = 0$ with

$$A = \begin{bmatrix} 1 & 1 & 1 \ 1 & -2 & 3 \ 1 & 4 & 9 \end{bmatrix}$$

Compute the determinant:

$$\det A = 1[(-2)(9) - (3)(4)] - 1[(1)(9) - (3)(1)] + 1[(1)(4) - (-2)(1)]$$

$$= 1(-18 - 12) - 1(9 - 3) + 1(4 + 2)$$

$$= -30 - 6 + 6 = -30 \neq 0$$

Since $\det A \neq 0$, the only solution is the trivial one.

Verification by row reduction

$$ \begin{bmatrix} 1 & 1 & 1 & 0 \ 1 & -2 & 3 & 0 \ 1 & 4 & 9 & 0 \end{bmatrix} \xrightarrow{R_2 - R_1,; R_3 - R_1} \begin{bmatrix} 1 & 1 & 1 & 0 \ 0 & -3 & 2 & 0 \ 0 & 3 & 8 & 0 \end{bmatrix} $$

$$ \xrightarrow{R_3 + R_2} \begin{bmatrix} 1 & 1 & 1 & 0 \ 0 & -3 & 2 & 0 \ 0 & 0 & 10 & 0 \end{bmatrix} $$

Back substitution:

$$10C_3 = 0 \implies C_3 = 0$$ $$-3C_2 + 2(0) = 0 \implies C_2 = 0$$ $$C_1 + 0 + 0 = 0 \implies C_1 = 0$$

Conclusion

The only solution is $C_1 = C_2 = C_3 = 0$ (trivial), so the signals $1^k$, $(-2)^k$, and $3^k$ are linearly independent. $\blacksquare$

(Note: this is expected since the three signals correspond to distinct roots $1, -2, 3$ of a characteristic equation, and modes with distinct roots are always independent.)