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Coordinate transformation. Transition matrix.

Let Ln− be an arbitrary dimensional space, B=(e1,...,en)− a fixed basis in it. Then each vector x∈Ln corresponds uniquely to the column of its coordinates in this basis.

x=x1e1+...+xnen⇔X=(x1⋮xn)

In this case, linear combinations over vectors in coordinate form look as follows:

z=x+y⇔Z=X+Y

y=λx⇔Y=λX.

Let B=(e1,e2,...,en) and B′=(e1′,e2′,...,en′)− be two different bases in Ln. Decompose each vector of the basis B′ into the basis B:

ek′=t1ke1+...+tnken⇔Ek′=(t1k ⋮ tnk),k=1,2,...,n.

The transition matrix TB→B′ from the basis B to the basis B′ is called the matrix

TB→B′=(t11...t1n.........tn1...tnn) the k-th column of which is the column Ek′ of coordinates of the vector ek′ in the basis B. If x− is an arbitrary vector from Ln, X and X′− the columns of its coordinates in the bases B and B′ respectively, then the equality holds X′=(TB→B′)−1X (the formula for coordinate transformation when changing the basis).

Examples.

4.15. In the space V3, the vectors e1′=i+j, e2′=i−j, e3′=−i+2j−k are given. Prove that the system B′=(e1′,e2′,e3′) is a basis in R3 and write the transition matrix TB→B′, where B=(e1=i,e2=j,e3=k). Find the coordinates of the vector x=i−2j+2k in the basis B′.

Solution.

To show that the system of vectors B′=(e1′,e2′,e3′) is a basis in R3, it is sufficient to show that these vectors are non-coplanar.

From the condition, we have e1′=i+j=(1,1,0), e2′=i−j=(1,−1,0), e3′=−i+2j−k=(−1,2,−1). The vectors e1′,e2′,e3′ are non-coplanar if |1101−10−12−1|≠0. Let's verify this:

|1101−10−12−1|=|−102−1|−|10−1−1|=1+1=2≠0. Therefore, the system B′=(e1′,e2′,e3′) is a basis in R3.

Next, we write the transition matrix TB→B′.

TB→B′=(t11...t1n.........tn1...tnn) the k-th column of which is the column Ek′ of coordinates of the vector ek′ in the basis B. That is, TB→B′=(11−11−1200−1).

Now, using the formula X′=(TB→B′)−1X, find the coordinates of the vector x=i−2j+2k in the basis B′. Here (TB→B′)−1=(11−11−1200−1)−1, X=(1−22).

Let's find the inverse matrix (TB→B′)−1:

Let A=TB→B′. Then detA=2;

A11=1; A12=1; A13=0;

A21=1; A22=−1; A23=0;

A31=1; A32=−3; A33=−2.

From here A∗=(1111−1−300−2); A−1=1detAA∗=12(1111−1−300−2).

Substituting this result into the formula X′=(TB→B′)−1X, we get:

X′=12(1111−1−300−2)(1−22)=12(1−3−4)=(1/2−3/2−2).

Answer: TB→B′=(11−11−1200−1); X′=(1/2−3/2−2).

4.17. Let B=(i,j,k) and B′=(i′,j′,k′) be orthogonal bases in R3. Write the transition matrix TB→B′, and list the column of coordinates for the vector x=i−2j+k in the basis B′.

The basis B′ is obtained by the permutation i′=j, j′=k, k′=i.

Solution.

From the condition, we have e1=i,e2=j,e3=k; e1′=j=(0,1,0), e2′=k=(0,0,1), e3′=i=(1,0,0).

Thus, the transition matrix is TB→B′=(001100010).

Now, using the formula X′=(TB→B′)−1X, let's find the coordinates of the vector x=i−2j+k in the basis B′. Here (TB→B′)−1=(001100010)−1, X=(1−21).

Let's find the inverse matrix (TB→B′)−1:

Designate A=TB→B′. Then detA=1;

A11=0; A12=0; A13=1;

A21=1; A22=0; A23=0;

A31=0; A32=1; A33=0.

From this, we have A∗=(010001100); A−1=1detAA∗=(010001100).

Substituting this result into the formula X′=(TB→B′)−1X, we obtain:

X′=(010001100)(1−21)=(−211).

Answer: TB→B′=(001100010); X′=(−211).

Homework:

Let B=(i,j,k) and B′=(i′,j′,k′) be orthogonal bases in R3. Write the transition matrix TB→B′, and list the column of coordinates for the vector x=i−2j+k in the basis B′.

4.16. The basis B′ is obtained by reversing the direction of all three basis vectors of B.

Answer: TB→B′=(−1000−1000−1); X′=(−12−1).

4.18. The basis B′ is obtained by rotating the basis B by an angle φ around the axis i.

Answer: TB→B′=(1000cos⁡φ−sin⁡φ0sin⁡φcos⁡φ); X′=(1−2cos⁡φ+sin⁡φ2sin⁡φ+cos⁡φ).

Tags: matrix, vector, coordinate transformation, transition matrix

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