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Given the basis vectors (v1, v2, v3) and (u1, u2, u3), find the transition matrix from the v-basis to the u-basis. If x = 2v1 + 3v2 − v3, determine the coordinates of x with respect to (u1, u2, u3).

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We are given two bases:
Basis B={v1⃗,v2⃗,v3⃗}B = \{\vec{v_1}, \vec{v_2}, \vec{v_3}\} where v1⃗=[407]\vec{v_1} = \begin{bmatrix} 4 \\ 0 \\ 7 \end{bmatrix}, v2⃗=[011]\vec{v_2} = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, v3⃗=[612]\vec{v_3} = \begin{bmatrix} 6 \\ 1 \\ 2 \end{bmatrix}.
Basis U={u1⃗,u2⃗,u3⃗}U = \{\vec{u_1}, \vec{u_2}, \vec{u_3}\} where u1⃗=[325]\vec{u_1} = \begin{bmatrix} 3 \\ 2 \\ 5 \end{bmatrix}, u2⃗=[012]\vec{u_2} = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}, u3⃗=[332]\vec{u_3} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}.

Part a· 3 steps

Find the transition matrix from (V1,V2,V3)(V_1, V_2, V_3) to (U1,U2,U3)(U_1, U_2, U_3).

  1. Form the augmented matrix $[U | V]$.

    The transition matrix PU←BP_{U \leftarrow B} is found by reducing the augmented matrix [U∣V][U | V] to [I∣PU←B][I | P_{U \leftarrow B}].

    [U∣V]=[303∣406213∣011522∣712][U | V] = \begin{bmatrix} 3 & 0 & 3 & | & 4 & 0 & 6 \\ 2 & 1 & 3 & | & 0 & 1 & 1 \\ 5 & 2 & 2 & | & 7 & 1 & 2 \end{bmatrix}

  2. Perform row operations to transform the left side into the identity matrix.

    R1→13R1R_1 \to \frac{1}{3}R_1:

    [101∣4/302213∣011522∣712]\begin{bmatrix} 1 & 0 & 1 & | & 4/3 & 0 & 2 \\ 2 & 1 & 3 & | & 0 & 1 & 1 \\ 5 & 2 & 2 & | & 7 & 1 & 2 \end{bmatrix}

    R2→R2−2R1R_2 \to R_2 - 2R_1 and R3→R3−5R1R_3 \to R_3 - 5R_1:

    [101∣4/302011∣−8/31−302−3∣1/31−8]\begin{bmatrix} 1 & 0 & 1 & | & 4/3 & 0 & 2 \\ 0 & 1 & 1 & | & -8/3 & 1 & -3 \\ 0 & 2 & -3 & | & 1/3 & 1 & -8 \end{bmatrix}

    R3→R3−2R2R_3 \to R_3 - 2R_2:

    [101∣4/302011∣−8/31−300−5∣17/3−1−2]\begin{bmatrix} 1 & 0 & 1 & | & 4/3 & 0 & 2 \\ 0 & 1 & 1 & | & -8/3 & 1 & -3 \\ 0 & 0 & -5 & | & 17/3 & -1 & -2 \end{bmatrix}

    R3→−15R3R_3 \to -\frac{1}{5}R_3:

    [101∣4/302011∣−8/31−3001∣−17/151/52/5]\begin{bmatrix} 1 & 0 & 1 & | & 4/3 & 0 & 2 \\ 0 & 1 & 1 & | & -8/3 & 1 & -3 \\ 0 & 0 & 1 & | & -17/15 & 1/5 & 2/5 \end{bmatrix}

    R1→R1−R3R_1 \to R_1 - R_3 and R2→R2−R3R_2 \to R_2 - R_3:

    [100∣37/15−1/58/5010∣−23/154/5−17/5001∣−17/151/52/5]\begin{bmatrix} 1 & 0 & 0 & | & 37/15 & -1/5 & 8/5 \\ 0 & 1 & 0 & | & -23/15 & 4/5 & -17/5 \\ 0 & 0 & 1 & | & -17/15 & 1/5 & 2/5 \end{bmatrix}

  3. Identify the transition matrix PU←BP_{U \leftarrow B}.

    The right side of the augmented matrix is the transition matrix.

    PU←B=[37/15−1/58/5−23/154/5−17/5−17/151/52/5]P_{U \leftarrow B} = \boxed{\begin{bmatrix} 37/15 & -1/5 & 8/5 \\ -23/15 & 4/5 & -17/5 \\ -17/15 & 1/5 & 2/5 \end{bmatrix}}

Part b· 3 steps

If x⃗=2v1⃗+3v2⃗−v3⃗\vec{x} = 2\vec{v_1} + 3\vec{v_2} - \vec{v_3}, determine the coordinates of x⃗\vec{x} with respect to UU.

  1. Write the coordinates of x⃗\vec{x} with respect to basis BB.

    Given x⃗=2v1⃗+3v2⃗−1v3⃗\vec{x} = 2\vec{v_1} + 3\vec{v_2} - 1\vec{v_3}, the coordinate vector of x⃗\vec{x} in basis BB is:

    [x⃗]B=[23−1][\vec{x}]_B = \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix}

  2. Use the transition matrix to find the coordinates of x⃗\vec{x} with respect to basis UU.

    The relationship is [x⃗]U=PU←B[x⃗]B[\vec{x}]_U = P_{U \leftarrow B} [\vec{x}]_B.

    [x⃗]U=[37/15−1/58/5−23/154/5−17/5−17/151/52/5][23−1][\vec{x}]_U = \begin{bmatrix} 37/15 & -1/5 & 8/5 \\ -23/15 & 4/5 & -17/5 \\ -17/15 & 1/5 & 2/5 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix}

  3. Perform the matrix multiplication.

    [x⃗]U=[(37/15)(2)+(−1/5)(3)+(8/5)(−1)(−23/15)(2)+(4/5)(3)+(−17/5)(−1)(−17/15)(2)+(1/5)(3)+(2/5)(−1)][\vec{x}]_U = \begin{bmatrix} (37/15)(2) + (-1/5)(3) + (8/5)(-1) \\ (-23/15)(2) + (4/5)(3) + (-17/5)(-1) \\ (-17/15)(2) + (1/5)(3) + (2/5)(-1) \end{bmatrix} [x⃗]U=[74/15−3/5−8/5−46/15+12/5+17/5−34/15+3/5−2/5][\vec{x}]_U = \begin{bmatrix} 74/15 - 3/5 - 8/5 \\ -46/15 + 12/5 + 17/5 \\ -34/15 + 3/5 - 2/5 \end{bmatrix}

    Convert to common denominator (15):

    [x⃗]U=[74/15−9/15−24/15−46/15+36/15+51/15−34/15+9/15−6/15][\vec{x}]_U = \begin{bmatrix} 74/15 - 9/15 - 24/15 \\ -46/15 + 36/15 + 51/15 \\ -34/15 + 9/15 - 6/15 \end{bmatrix} [x⃗]U=[(74−9−24)/15(−46+36+51)/15(−34+9−6)/15][\vec{x}]_U = \begin{bmatrix} (74 - 9 - 24)/15 \\ (-46 + 36 + 51)/15 \\ (-34 + 9 - 6)/15 \end{bmatrix} [x⃗]U=[41/1541/15−31/15][\vec{x}]_U = \begin{bmatrix} 41/15 \\ 41/15 \\ -31/15 \end{bmatrix}

    The coordinates of x⃗\vec{x} with respect to basis UU are:

    [x⃗]U=[41/1541/15−31/15][\vec{x}]_U = \boxed{\begin{bmatrix} 41/15 \\ 41/15 \\ -31/15 \end{bmatrix}}
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