Linear Algebra Midterm 3

Determine if the set of vectors is orthonormal.

Find inner product of vectors and make sure it = 0. See if they are unit vectors. Basically u • u = 1 or v • v = 1

Let y, u1 and u2 be vectors in ℝ^3. Find the distance from y to the plane in ℝ^3 spanned by u1 and u2.

Find y hat using ŷ=(y•uj/uj•uj) Then find y-ŷ. Then square the numbers in y-ŷ and square root their sums.

How can the characteristic polynomial of a matrix A be​ determined?

Find ​det(A-λ​I), where λ is a scalar.

For a system of linear equations Xβ=y​, what is the design matrix​ X?

First column is all 1s, 2nd column are the x values of the coordinates.

What does it mean if A is​ diagonalizable?

If A is​ diagonalizable, then AequalsPDP Superscript negative 1 for some invertible P and diagonal D.

Which theorem could help prove one of these criteria necessary for a set of vectors to be an orthogonal basis for a subspace W of set of real numbers ℝ^n​ from​ another?

If S={u1, . . . up} is an orthogonal set of nonzero vectors in ℝ^n​, then S is linearly independent and hence is a basis for the subspace spanned by S.

Determine whether the given matrix is regular. Explain your answer. P= [ 2/7 0 4/5 0 ] [ 0 2/3 0 1/2 ] [ 5/7 0 1/5 0 ] [ 0 1/3 0 1/2 ]

It is not regular. There is no power of P that contains only strictly positive entries.

Determine whether the given matrix is regular. Explain your answer. [ 1/3 0 1/2 ] P = [ 1/3 1/2 1/2 ] [ 1/3 1/2 0 ]

It is regular. The first power of P that contains only strictly positive entries is P^2/

Since X^TX is ________, any equation X^TXx = b, such as the normal equation __________ has a unique solution according to the __________. This solution is the​ least-squares solution to the system Xβ=y. So, the normal equations have a unique solution if and only if the data include at least two data points with different​ x-coordinates.

Since X^TX is invertible, any equation X^TXx = b, such as the normal equation X^TXβ = X^Ty has a unique solution according to the Invertible Matrix Theorem. This solution is the​ least-squares solution to the system Xβ=y. So, the normal equations have a unique solution if and only if the data include at least two data points with different​ x-coordinates.

If A is​ diagonalizable, then A has n distinct eigenvalues.

The statement is false. A diagonalizable matrix can have fewer than n eigenvalues and still have n linearly independent eigenvectors.

A matrix A is diagonalizable if A has n eigenvectors.

The statement is false. A diagonalizable matrix must have n linearly independent eigenvectors.

If A is​ invertible, then A is diagonalizable.

The statement is false. An invertible matrix may have fewer than n linearly independent​ eigenvectors, making it not diagonalizable.

If z is orthogonal to u1 and to u2 and if W = Span{u1, u2}, then z must be in W⊥.

This statement is true because, since z is orthogonal to every vector in span {u1, u2}, a set that spans W.

The least squares solution, β^, to Xβ=y​ is the unique solution to . . .

X^TXβ = X^Ty

Why does this show that a square matrix U with orthonormal columns is​ invertible?

A square matrix is invertible if and only if its columns are linearly independent.

What is the inverse of​ A?

A^-1=PD^-1P^-1

If the distance from u to v equals the distance from u to minus​v, then u and v are orthogonal

By the definition of​ orthogonal, u and v are orthogonal if and only if u • v = 0. This happens if and only if 2u•v = -2u•v, which happens if and only if the squared distance from u to v equals the squared distance from u to -v. Requiring the squared distances to be equal is the same as requiring the distances to be​ equal, so the given statement is true.

Compute the distance from y to the line through u and the origin. Where y and u are vectors in |R^2

Calculate ŷ by cj=(y•uj/uj•uj). Then calculate y - ŷ. Once you get that vector. Square the numbers and square root the sum to find distance.

A least-squares solution is a solution to Ax=b.

False

B̂ is the least-squares solution of Ax=b

False

A matrix with orthonormal columns is an orthogonal matrix.

False. A matrix with orthonormal columns is an orthogonal matrix if the matrix is also square.

If {v1, v2, v3} is an orthogonal basis for W​,then multiplying v3 by a scalar c gives a new orthogonal basis {v1, v2, cv3}.

False. If the scale factor is zero it will not give a new orthogonal basis.

If the vectors in an orthogonal set of nonzero vectors are​ normalized, then some of the new vectors may not be orthogonal.

False. Normalization changes all nonzero vectors to have unit​ length, but does not change their relative angles.​ Therefore, orthogonal vectors will always remain orthogonal after they are normalized.

If L is a line through 0 and if ŷ is the orthogonal projection of y onto​ L, then ||ŷ|| gives the distance from y to L.

False. The distance from y to L is given by ||y - ŷ||.

A is a 3x3 matrix with two eigenvalues. Each eigenspace is​ one-dimensional. Is A​ diagonalizable? Why?

No. The sum of the dimensions of the eigenspaces equals 2 and the matrix has 3 columns. The sum of the dimensions of the eigenspace and the number of columns must be equal.

Why is it true that the columns of U are linearly​ independent?

Orthonormal sets are linearly independent sets.

Verify that {u1, u2} is an orthogonal​ set, and then find the orthogonal projection of y onto {u1, u2}.

See if u1 and u2 are orthogonal. Use ŷ=(y•uj/uj•uj) for each vector and add them together.

How do these calculations show that {u1, u2, u3} is an orthogonal basis for ℝ^3​?

Since each inner product is 0​, the vectors form an orthogonal set. From the theorem​ above, this proves that the vectors are also a basis.

Given the property from the previous​ step, which of the following are equivalent properties to that​ property?

The equation Xβ=y​ has a unique​ least-squares solution for each y in ℝ^m AND the matrix X^TX is invertible

For a square matrix ​A, vectors in ColA are orthogonal to vectors in NulA.

The given statement is false. By the theorem of orthogonal​ complements, it is known that vectors in ColA are orthogonal to vectors in NulA^T. Using the definition of orthogonal​ complements, vectors in ColA are orthogonal to vectors in NulA if and only if the rows and columns of A are the​ same, which is not necessarily true.

For any scalar c, ||cv|| = c||v||

The given statement is false. Since length is always​ positive, the value of ||cv|| will always be positive. By the same​ logic, when c is​ negative, the value of c||v|| is negative.

For any scalar c, u • (cv)=c (u • v)

The given statement is true because this is a valid property of the inner product.

If ||u||^2 + ||v||^2 = ||u+v||​^2, then u and v are orthogonal.

The given statement is true. By the Pythagorean​ Theorem, two vectors u and v are orthogonal if and only if||u||^2 + ||v||^2 = ||u+v||​^2.

v • v = ||v||^2

The given statement is true. By the definition of the length of a vector v​, ||v||=√v•v

For an mxn matrix​ A, vectors in the null space of A are orthogonal to vectors in the row space of A.

The given statement is true. By the theorem of orthogonal complements, (RowA)^⊥ = Nul A. It follow , by the definition of orthogonal complements, that vectors in the null space of A are orthogonal to the row space of A

If vectors v1​,...,vp span a subspace W and if x is orthogonal to each vj j for j=1,...,p, then x is in W^⊥.

The given statement is true. If x is orthogonal to each vj​, then x is also orthogonal to any linear combination of those vj. Since any vector in W can be described as a linear combination of vj​, x is orthogonal to all vectors in W.

If x is orthogonal to every vector in a subspace​ W, then x is in W^⊥.

The given statement is true. If x is orthogonal to every vector in​ W, then x is said to be orthogonal to W. The set of all vectors x that are orthogonal to W^⊥

u • v - v • u

The given statement is true. Since the inner product is​ commutative, u • v = v • u. Subtracting v•u from each side of this equation gives u•v-v•u=0.

Given a 2x2 matrix​ A, where Upper A equals PCP^-1 and C is of the form [ a -b ] [ b a ]​, what is the form of​ P?

The matrix [ Rev lmv ], where A has a complex eigenvalue λ=a-bi ​(b/=0) and an associated eigenvector v in set of complex numbers C^2

Note that the distance from a point y in ℝ^3 to a subspace W is defined as the distance from y to the closest point in W. What is the closest point in W to y​?

The orthogonal projection of y onto W

A​ least-squares solution of Ax=b is a vector x^ such that ||b - Ax|| ≤ ||b - Ax^||for all x in |R^n

The statement is false because a​ least-squares solution of Ax=b is a vector x^such that ||b-Ax^|| ≤ ||b - Ax|| for all x in |R^n

If A is​ diagonalizable, then A is invertible.

The statement is false because invertibility depends on 0 not being an eigenvalue. A diagonalizable matrix may or may not have 0 as an eigenvalue.

A is diagonalizable if and only if A has n​ eigenvalues, counting multiplicities.

The statement is false because the eigenvalues of A may not produce enough eigenvectors to form a basis of |R^n.

A is diagonalizable ifA=PDP^-1 for some matrix D and some invertible matrix P.

The statement is false because the symbol D does not automatically denote a diagonal matrix.

The orthogonal projection of ŷ of y onto a subspace W can sometimes depend on the orthogonal basis for W used to compute ŷ.

The statement is false because the uniqueness property of the orthogonal decomposition y = ŷ+z indicates​ that, no matter the basis used to find​ it, it will always be the same.

If |R^n has a basis of eigenvectors of​ A, then A is diagonalizable.

The statement is true because A is diagonalizable if and only if there are enough eigenvectors to form a basis of |R^n.

A​ least-squares solution of Ax=b is a vector x^ that satisfies Ax^=b^where b^ is the orthogonal projection of b onto Col A.

The statement is true because b^ is the closest point in Col A to b. So, Ax=b^ is consistent and x^such that Ax^=b^ is a least squares solution from Ax=b.

If y is in a subspace W, then the orthogonal projection of y onto W is y itself.

The statement is true because for an orthogonal basis of​ W, B= u1, ...up, y and proj(w)y can be written as linear combinations of vectors in B with equal weights.

If the columns of A are linearly​ independent, then the equation Ax=b has exactly one​ least-squares solution.

The statement is true because if the columns of A are linearly​ independent, then A^TA is invertible and x^=(A^TA)^-1 A^Tb is the least squared solution to Ax=b.

If the columns of an nxp matrix U are orthonormal, then UU^Ty is the orthogonal projection of y onto the column space of U.

The statement is true because the columns of U are linearly independent and thus form an orthonormal basis for Col U.​ So, proj(ColU)y = UU^Tyfor all y in R^n

The general​ least-squares problem is to find an x that makes Ax as close as possible to b.

The statement is true because the general​ least-squares problem attempts to find an x that minimizes ||b −Ax||.

Any solution of A^TAx = A^T b is a least squared solution of Ax=b.

The statement is true because the set of​ least-squares solutions of Ax=b coincides with the nonempty set of solutions of the normal​ equations, defined as A^TAx = A^T b

For each y and each subspace W , the vector y − proj(w) y is orthogonal to W.

The statement is true because y can be uniquely written in the form y = proj(w)y +z where proj(w)y is in W and z is in W⊥ and it follows that z =y-proj(w)y

If AP=​PD, with D​ diagonal, then the nonzero columns of P must be eigenvectors of A.

The statement is true. Let v be a nonzero column in P and let lambda be the corresponding diagonal element in D. Then AP=PD implies that Av=λv​, which means that v is an eigenvector of A.

Write y as the sum of a vector in Span {u} and a vector orthogonal to u.

The vector y can be written as the sum of two orthogonal​ vectors, one in Span {u}​, ŷ, and one orthogonal to u, y - ŷ.

Find the best approximation to z by vectors of the form c1v1+c2v2. Let W be a subspace of ℝ^n​, let y be any vector in ℝ^n​,and let ŷ be the orthogonal projection of y onto W.

Then ŷ is the closest point in W to y in the sense that ||y - ŷ|| < ||y-v|| for all v distinct from ŷ.

Let U be a square matrix with orthonormal columns. Explain why U is invertible. Which of the following is true of the columns of​ U?

They are linearly independent. The inner product of each pair of vectors is 0. Each column vector has unit length of 1.

B̂ is in the closest point in Col A to b

True

B̂ is in the column space of A

True

The least squares solution of Ax = b is a vector, x, that minimizes the distance between b and Ax.

True

The vector Ax is in the column space of A.

True

Thus, least squares solutions of Xβ=y coincides with the nonempty set of solutions of the normal equations of X^TXβ = X^Ty

True

The​ Gram-Schmidt process produces from a linearly independent set {x1, . . ., xp} an orthogonal set {v1, . . ., vp} with the property that for each k, the vectors v1, . . ., vk span the same subspace as that spanned by x1, . . ., xk.

True. For Wk=​Span {x1, . . ., xk}, v1=x1 and some v1, . . ., vk where {v1, . . ., vk} is an orthogonal basis for Wk, v k+1=x k+1 - proj(Wk) x k+1is orthogonal to Wk and is in W k+1. Also v k+1 /=0. Hence, {v1, . . ., vk} is an orthogonal basis for W k+1.

If y is a linear combination of nonzero vectors from an orthogonal​ set, then the weights in the linear combination can be computed without row operations on a matrix.

True. For each y in​ W, the weights in the linear combination y=c1u1+...+cpup can be computed by cj=(y•uj/uj•uj) where j= 1, . . ., p.

Not every linearly independent set in ℝ^n is an orthogonal set.

True. For​ example, the vectors [0] [1] [1] [1] are linearly independent but not orthogonal.

If A=​QR, where Q has orthonormal​ columns, then R=Q^TA.

True. Since Q has orthonormal columns then Q^TQ=I. So Q^TA=QT(QR)=IR=R.

Which of the following criteria are necessary for a set of vectors to be an orthogonal basis for a subspace W of set of real numbers ℝ^n​? Select all that apply.

Vectors must span W. Vectors must form an orthogonal set.

What are the normal equations that should be used to solve for β​?

X^TXβ​ = X^Ty

What does it mean if A is​ invertible?

Zero is not an eigenvalue of​ A, so the diagonal entries in D are not​ zero, so D is invertible.

Identify a nonzero 2times2 matrix that is invertible but not diagonalizable.

[ 1 1 ] [ 1 0 ] [ 0 1 ] not [ 0 1 ]

If at least two data points have different​ x-coordinates, the columns of the design matrix . . .

are linearly independent

Vectors u and v are orthogonal if . . .

u • v = 0