1. Let P = (1, 2, 1) and let R be the line {(1, 5, −1) + t(1, 1, −1)}. Compute the distance between P and L and find the point of L which is nearest to P .

Linear Algebra and Geometry. Written test of september 12, 2013.

1. Let P = (1, 2, 1) and let R be the line {(1, 5, −1) + t(1, 1, −1)}. Compute the distance between P and L and find the point of L which is nearest to P .

Solution. Let u = (1, 2, 1) − (1, 5, −1) = (0, −3, 2). We have that the projection of u along L((1, 1, −1) is pr(u) = −5/3(1, 1, −1). Therefore the nearest point is

H = (1, 5, −1) − 5/3(1, 1, −1) The distance is

k u − pr(u) k=k (0, −3, 2) + 5/3(1, 1, −1) k=k 1/3(5, −4, −1) k= √ 42/3

2. A particle moves along the ellipse 2x ² + y ² = 1 with position vector r(t) = (x(t), y(t)).

The motion is such that x ⁰ (t) = −4y(t) for each t.

(a) Prove that y ⁰ (t) is proportional to x(t) and find the factor of proportionality;

(b) How much time is required for the particle to go once around the ellipse?

Solution. We have that 2x(t) ² + y(t) ² = 1 for each t. Differentiating we get that 4x(t)x ⁰ (t) + 2y(t)y ⁰ (t) = 0. Since x ⁰ (t) = −4y(t) we get −16x(t)y(t) + 2(y(t)y ⁰ (t) = 2y(t)(−8x(t) + y ⁰ (t)) ≡ 0. Since y(t) is non-zero except for some isolated points of t we conclude, from the continuity of y ⁰ (t), that y ⁰ (t) = 8x(t) for each t.

(b) We have that (x ⁰ (t), y ⁰ (t) = (−4y(t), 8x(t)) Differentiating another time we get

(x ⁰⁰ (t), y ⁰⁰ (t)) = (−32x(t), −32y(t)). It turns out that both x(t) and y(t) are solutions of the equation f ⁰⁰ = −32f . Therefore they are of the form a sin 4 √

2t + b cos 4 √

2t. Hence the time to go once around the ellipse is T = 2π/(4 √

2).

3. In V 4 , with the usual dot product, let W = L((1, 1, −1, 0), (1, 0, 0, 1)) and let v = (1, 1, 1, 0).

(a) Find a vector w ∈ W and a vector u ∈ W ^⊥ such that v = w + u.

(b) Is it possible to find two other vectors w ⁰ ∈ W and u ⁰ ∈ W ^⊥ , different from w and u, such that v = w ⁰ + u ⁰ ?

Solution. (b) By the Theorem on Orthogonal Decompositions, vectors w and u as required do exist, and they are unique. Therefore the answer to (b) is NO.

(a) By the Theorem on Orthogonal Decompositions, we have that w must be the projection of v onto W , which will be denoted pr _W (v) , and u = v − pr _W (v). Therefore we have only to compute the projection of v onto W . In order to do that we have first to orthogonalize the given basis of W . We take w ₁ = (1, 1, −1, 0). Moreover we have that

(1, 0, 0, 1) − pr _L(v

₎ ((1, 0, 0, 1)) = (1, 0, 0, 1) − (1/3)(1, 1, −1, 0) = (1/3)(2, −1, 1, 3) Therefore, to simplify the calculation, we can take w 2 = (2, −1, 1, 3). Finally, se have that

pr _W (v) = pr _L(w

1

) (v) + pr _L(w

2

) (v) = (1/3)(1, 1, −1, 0) + (6/5)(2, −1, 1, 3)

1

As already mentioned before u := v − pr _W (v) belongs to W ^⊥ .

4. Let T : V 3 → V 3 be the linear transformation such that T ((1, 0, 1)) = (1, 1, 2), T ((1, 0, −1) = (0, 2, 3) and T ((1, 2, −1)) = (1, 3, 5).

(a) Find the dimension and a basis of the range T (V 3 );

(b) find the dimension and a basis of the null-space N (T );

(c) find the matrix of T with respect to the canonical basis of V 3 .

Solution. (a) We denote v ₁ = (1, 0, 1), v ₂ = (1, 0, −1) and v ₃ = (1, 2, −1). Since they are independent (easily seen), they form a basis B of V 3 . An easy computation shows that T (v ₁ ), T (v ₂ ) and T (v ₃ ) are not independent. Since the first two are independent, it follows that rk(T ) = 2 and that {(1, 1, 2), (0, 2, 3)} is a basis of T (V 3 ).

(b) By the nullity + rank Theorem, the nullity is 1. To find a basis, the simpler way is to note that T (v 1 ) + T (v 2 ) = T (v 3 ). This means that T (v 1 + v 2 − v 3 ) = 0, that is v ₁ + v ₂ − v ₃ ∈ N (T ). Since we know that dim N (T ) = 1, we conclude that {v ₁ + v ₂ − v ₃ } is a basis of N (T ).

If one does not see that T (v 1 ) + T (v 2 ) = T (v 3 ), one can argue as follows. We have that

B = m ^B _E (T ) =





1 0 1 1 2 3 2 3 5





The solutions of the system B



 x ⁰ y ⁰ z ⁰



 =



 0 0 0



 are the components – with respect to the basis B – of vectors v such that T (v) = 0. It is easily seen that the solutions of the above system are the triples (x ⁰ , y ⁰ z ⁰ ) such that x ⁰ = −z ⁰ and y ⁰ = −z ⁰ , that is of the form λ(1, 1, −1) for λ ∈ R. We conclude, as above, that N (T ) is spanned by 1 v 1 + 1 v 2 + (−1) v 3 .

(c) We have that

m ^E _E (T ) = m ^B _E (T ) m ^E _B (id)

(you can see this by writing down the corresponding diagram of linear transformations).

Since m ^E _B (id) = m ^B _E (id) ⁻¹ = C ⁻¹ , where

C =





1 1 1

0 0 2

1 −1 −1





we finally have

m ^E _E (T ) = B C ⁻¹

5. (a) Find all real 3 × 3 symmetric matrices A such that: 2 is a double eigenvalue of A, det A = −12 and L((1, 1, 1)) is an eigenspace of A.

(b) For a matrix A as in (a), let us consider the function Q A : V 3 → R defined by

2

Q _A (X) = X A X ^t . Find a triple ¯ X = (¯ x ₁ , ¯ x ₂ , ¯ x ₃ ) such that Q _A ( ¯ X) is positive and a triple Y = (¯ ¯ y 1 , ¯ y 2 , ¯ y 3 ) such that Q A ( ¯ Y ) is negative.

Solution. (a) The matrix A must have another (simple) eigenvalue. Since the product of (possibly coinciding) eigenvalues of A equals det A = −12, this simple eigenvalue is

−3. Since A is symmetric, it is diagonalizable, hence the eigenspace corresponding to the double eigenvalue must have dimension 2. Therefore L((1, 1, 1)) must be the eigenspace E(−3). Since A is symmetric, the eigenspace E(2) must be E(−3) ^⊥ , hence it is defined by the equation x + y + z = 0. Hence E(−3) = L((−1, 1, 0), (−1, 0, 1)). Putting together the basis of E(2) and the basis of E(−3) we get a basis B of V ₃ . We consider the matrix C = m ^B _E (id) =





−1 −1 1

1 0 1

0 1 1



. We know that diag(2, 2, −3) = C ⁻¹ A C. Therefore the final answer to A is the following: there is one and only one such matrix A, namely

A = C diag(2, 2, −3) C ⁻¹

(b) Note that the function Q A is the quadratic form associated to the matrix A. Certainly an eigenvector ¯ X of 2 – for example ¯ X = (−1, 1, 0) – is such that Q _A ( ¯ X) > 0. Indeed Q A ( ¯ X) = ¯ X A ¯ X ^t = ¯ X 2 ¯ X ^t = 2 ¯ X · ¯ X > 0. By the same reason, an eigenvector ¯ Y of −3, for example (1, 1, 1), is such that Q( ¯ Y ) < 0.

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