Case 3: Let E be obtained by adding a multiple of one row of In to another. If either A or B is singular, then either det A or det B is zero. Thus AB is also singular. If it could, then it would be invertible as the product of invertible matrices. The reduced row echelon form of A is the product of A and elementary matrices, all of which are invertible. In general, reversing the order of the columns may change the sign of the determinant. There are 24 terms in this sum. Since the product of integers is always an integer, each elementary product is an integer.
The result then follows from the fact that the sum of integers is always an integer. Now consider any elementary product a1j1a2j2 … anjn. Hence, a11a22 … ann is the only elementary product which is not guaranteed to be zero. Since the column indices in this product are in natural order, the product appears with a plus sign. Thus, the determinant of U is the product of its diagonal elements. A similar argument works for lower triangular matrices.
See Theorem 2. We simply expand W. This will insure that the sum of the products of corresponding entries from the ith row of A and the ith column of A—1 will remain equal to 1. Call that matrix B. Now suppose that we add —c times the jth column of A—1 to the ith column of A—1.
Call that matrix C. Supplementary Exercises 2 85 It is less obvious that c is the trace of the matrix of minors of the entries of A; that is, the sum of the minors of the diagonal entries of A.
If we multiply Column 1 by , Column 2 by , Column 3 by , Column 4 by 10, and add the results to Column 5, we obtain a new Column 5 whose entries are just the 5 numbers listed in the problem. Since each is divisible by 19, so is the resulting determinant. Exercise Set 3.
Suppose there are scalars c1, c2, and c3 which satisfy the given equation. Clearly, there do not exist scalars c1, c2, and c3 which satisfy the above equation, and hence the system is inconsistent. Let X be a point on the line through P and Q and let t PQ where t is a positive, real number be the vector with initial point P and terminal point X. The vector u has terminal point Q which is the midpoint of the line segment connecting P1 and P2.
These proofs are for vectors in 3-space. To obtain proofs in 2-space, just delete the 3rd component. See Exercise 9. Equality occurs only when u and v have the same direction or when one is the zero vector. Thus the vectors are orthogonal. This is just the Pythagorean Theorem. Choose any nonzero vector w which is not parallel to u.
Now assign nonzero values to any two of the variables x, y, and z and solve for the remaining variable. By Theorem 3. But also by Theorem 3. Hence, their cross-product is zero. Hence, it must lie on the line through the origin perpendicular to v and in the plane determined by u and v.
Applying Part d of Theorem 3. Since these vectors are not multiples of one another, the planes are not parallel. Since one vector is three times the other, the planes are parallel.
Since the inner product of these two vectors is not zero, the planes are not perpendicular. Alternatively, recall that a direction vector for the line is just the cross-product of the normal vectors for the two planes, i. Since the plane is perpendicular to a line with direction 2, 3, —5 , we can use that vector as a normal to the plane. Call the points A, B, C, and D, respectively. Since they have points in common, they must coincide. A normal n to a plane which is perpendicular to both of the given planes must be perpendicular to both n1 and n2.
These, together with the given point and the methods of Example 2, will yield an equation for the desired plane. We change the parameter in the equations for the second line from t to s. Hence, the two lines coincide. They both pass through the point r0 and both are parallel to v. This represents a line through the point P1 with direction r2 — r1. Hence the given equation represents this line segment. Thus the system is inconsistent, so the lines are skew. Exercise Set 4. The transformation is not linear because of the terms 2x1x2 and 3x1x2.
In matrix terms, a dilation or contraction is represented by a scalar multiple of the identity matrix. Since such a matrix commutes with any square matrix of the appropriate size, the transformations commute.
Compute the trace of the matrix given in Formula 17 and use the fact that a, b, c is a unit vector. Thus 3, 1 , for example, is not in the range. Thus S is not linear. Thus, the Lagrange expression must be algebraically equivalent to the Vandermonde form.
This is done by adding a next term to p i , pe. This is a vector space. We shall check only four of the axioms because the others follow easily from various properties of the real numbers. The details are easily checked. Let k and m be scalars. Axiom 4: There is no zero vector in this set. Thus, there is no one zero vector that will work for every vector a, b, c in R3.
Since we are using standard matrix addition and scalar multiplication, the majority of axioms hold. However, the following axioms fail for this set V: Axiom 1: Clearly if A is invertible, then so is —A. Thus, V is not a vector space. Since we are using the standard operations of addition and scalar multiplication, Axioms 2, 3, 5, 7, 8, 9, 10 will hold automatically.
However, for Axiom 4 to hold, we need the zero vector 0, 0 to be in V. Thus, the set of all points in R2 lying on a line is a vector space exactly in the case when the line passes through the origin. Exercise Set 5. However, for Axiom 4 to hold, we need the zero vector 0, 0, 0 to be in V.
Thus, the set of all points in R3 lying on a plane is a vector space exactly in the case when the plane passes through the origin. Planes which do not pass through the origin do not contain the zero vector.
Since this space has only one element, it would have to be the zero vector. In fact, this is just the zero vector space. Suppose that u has two negatives, —u 1 and —u 2. We have proved that it has at most one. Thus it is not a subspace.
Therefore, it is a subspace of R3. The same is true of a constant multiple of such a polynomial. Hence, this set is a subspace of P3. Hence, the subset is closed under vector addition. Thus, the subset is not closed under scalar multiplication and is therefore not a subspace. Thus the set is a subspace. Thus 2, 2, 2 is a linear combination of u and v. Thus, the system of equations is inconsistent and therefore 0, 4, 5 is not a linear combination of u and v.
Since the determinant of the system is nonzero, the system of equations must have a solution for any values of x, y, and z, whatsoever. Therefore, v1, v2, and v3 do indeed span R3.
Note that we can also show that the system of equations has a solution by solving for a, b, and c explicitly. Since this is not the case for all values of x, y, and z, the given vectors do not span R3.
Hence the given polynomials do not span P2. The set of solution vectors of such a system does not contain the zero vector. Hence it cannot be a subspace of Rn. Alternatively, we could show that it is not closed under scalar multiplication. Let u and v be vectors in W. Let W1 and W2 be subspaces of V. This follows from the closure of both W1 and W2 under vector addition and scalar multiplication.
They cannot all lie in the same plane. Hence, the four vectors are linearly independent. This implies that k3 and hence k2 must also equal zero. Thus the three vectors are linearly independent. Thus they do not lie in the same plane. Suppose that S has a linearly dependent subset T. Denote its vectors by w1,…, wm. Since not all of the constants are zero, it follows that S is not a linearly independent set of vectors, contrary to the hypothesis. That is, if S is a linearly independent set, then so is every non-empty subset T.
This is similar to Problem Use Theorem 5. The set has the correct number of vectors. Thus the desired coordinate vector is 3, —2, 1. For instance, 1, —1, —1 and 0, 5, 2 are a basis because they satisfy the plane equation and neither is a multiple of the other. For instance, 2, —1, 4 will work, as will any nonzero multiple of this vector. Hence it is a basis for P2.
There is. Thus, its solution space should have dimension n — 1. Since AT is also invertible, it is row equivalent to In. It is clear that the column vectors of In are linearly independent.
Hence, by virtue of Theorem 5. Therefore the rows of A form a set of n linearly independent vectors in Rn, and consequently form a basis for Rn. Any invertible matrix will satisfy this condition. The nullspace of D is the entire xy-plane. Use Theorems 5. However A must be the zero matrix, so the system gives no information at all about its solution. That is, the row and column spaces of A have dimension 2, so neither space can be a line. Rank A can never be 1. Thus, by Theorem 5.
Hence, Theorem 5. Verify that these polynomials form a basis for P 1. Exercise Set 6. To prove Part a of Theorem 6. To prove Part d , observe that, by Theorem 5. By inspection, a normal vector to the plane is 1, —2, —3.
From the reduced form, we see that the nullspace consists of all vectors of the form 16, 19, 1 t, so that the vector 16, 19, 1 is a basis for this space. Conversely, if a vector w of V is orthogonal to each basis vector of W, then, by Problem 20, it is orthogonal to every vector in W. In fact V is a subspace of W c True. The two spaces are orthogonal complements and the only vector orthogonal to itself is the zero vector.
For instance, if A is invertible, then both its row space and its column space are all of Rn. See Exercise 3, Parts b and c. The set is therefore orthogonal. It will be an orthonormal basis provided that the three vectors are linearly independent, which is guaranteed by Theorem 6. Note that u1 and u2 are orthonormal. Thus we apply Theorem 6. By Theorem 6. But v1 is a multiple of u1 while v2 is a linear combination of u1 and u2.
This is similar to Exercise 29 except that the lower limit of integration is changed from —1 to 0. Then if u is any vector in V, we know from Theorem 6. Moreover, this decomposition of u is unique. Theorem 6. If the vectors vi form an orthogonal set, not necessarily orthonormal, then we must normalize them to obtain Part b of the theorem.
However, although they are orthogonal with respect to the Euclidean inner product, they are not orthonormal. However, they are neither orthogonal nor of unit length with respect to the Euclidean inner product. Suppose that v1, v2, …, vn is an orthonormal set of vectors.
Thus, the orthonormal set of vectors cannot be linearly dependent. The zero vector space has no basis 0. This vector cannot be linearly independent. If A is a necessarily square matrix with a nonzero determinant, then A has linearly independent column vectors.
Thus, by Theorem 6. Hence the error vector is orthogonal to the column space of A. Therefore Ax — b is orthogonal to the column space of A. Since the row vectors and the column vectors of the given matrix are orthogonal, the matrix will be orthogonal provided these vectors have norm 1.
Note that A is orthogonal if and only if AT is orthogonal. Since the rows of AT are the columns of A, we need only apply the equivalence of Parts a and b to AT to obtain the equivalence of Parts a and c. If A is the standard matrix associated with a rigid transformation, then Theorem 6. But if A is orthogonal, then Theorem 6.
Exercise Set 7. By Theorem 7. Thus by Theorem 7. Since A has no real eigenvalues, there are no lines which are invariant under A. Let aij denote the ijth entry of A. Since the eigenvalues are assumed to be real numbers, the result follows.
The converse of this result is also true. This is a straightforward computation and we leave it to you. The matrices 3I — A and 2I — A both have rank 2 and hence nullity 1.
Thus A has only 2 linearly independent eigenvectors, so it is not diagonalizable. Any matrix Q which is obtained from P by multiplying each entry by a nonzero number k will also work.
Suppose that A is invertible and diagonalizable. In addition, Theorem 7. In other words, Dk displays the eigenvalues of Ak along its diagonal. The sequence diverges for all other values of a. Thus each eigenvalue is repeated once and hence each eigenspace is 1-dimensional. By the result of Exercise 17, Section 7. Moreover, if A has nonnegative eigenvalues, then the diagonal entries of D are nonnegative since they are all eigenvalues.
But there is no elementary product which contains exactly n — 1 of these factors Why? Supplementary Exercises 7 7. We know that A has at most n eigenvalues, so that this expression can take on only finitely many values. Thus the only possible eigenvalues of A are zero and tr A. It is easy to check that each of these is, in fact, an eigenvalue of A. Since every odd power of A is again A, we have that every odd power of an eigenvalue of A is again an eigenvalue of A.
This says that T leaves every point in the x-y plane unchanged. All linear transformations have this property, and, for instance there is more than one linear transformation from R2 to R2. But there is only one linear transformation which maps every vector to the zero vector. Exercise Set 8. Theorem 5. Hence 5, 0 is not in R T. By Theorem 8. Since the only subspaces of R3 are the origin, a line through the origin, a plane through the origin, or R3 itself, the result follows.
It is clear that all of these possibilities can actually occur. These are parametric equations for a line through the origin. That range, which we can interpret as a subspace of R3, is a plane through the origin.
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