Linear Algebra Quiz 424
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 utimes•vequals=00. This happens if and only if 2 Bold u times Bold v equals negative 2 Bold u times Bold v comma2u•v=−2u•v, which happens if and only if the squared distance from u to v equals the squared distance from u to minus−v. Requiring the squared distances to be equal is the same as requiring the distances to be equal, so the given statement is true.
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 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 ModifyingAbove Bold y with carety is the orthogonal projection of y onto L, then left norm ModifyingAbove Bold y with caret right normy gives the distance from y to L.
False. The distance from y to L is given by left norm y minus ModifyingAbove Bold y with caret right normy−y.
If a set Sequals=StartSet Bold u 1 comma . . . comma Bold u Subscript p EndSetu1, . . . , up has the property that Bold u Subscript i Baseline times Bold u Subscript j Baseline equals 0ui•uj=0 whenever i not equals ji≠j, then S is an orthonormal set.
False. To be orthonormal, the vectors in S must be unit vectors as well as being orthogonal to each other.
How can this inverse be expressed using transposes? Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
Since U and V are orthogonal matrices, Upper U Superscript negative 1U−1equals=Upper U Superscript Upper TUT and Upper V Superscript negative 1V−1equals=Upper V Superscript Upper TVT. By substitution, left parenthesis UV right parenthesis Superscript negative 1(UV)−1equals=Upper V Superscript Upper T Baseline Upper U Superscript Upper TVTUT.
Why is UV invertible?
Since U and V are orthogonal, each is invertible by the definition of orthogonal matrices. The product of two invertible matrices is also invertible.
For an mtimes×n 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, left parenthesis Row font size decreased by 3 Upper A right parenthesis Superscript orthogonal(Row A)⊥equals=Nul font size decreased by 3 A.Nul A. It follows, by the definition of orthogonal complements, that vectors in the null space of A are orthogonal to vectors in the row space of A.
For a square matrix A, vectors in Co l font size decreased by 3Col A are orthogonal to vectors in Nu l font size decreased by 3Nul A.
The given statement is false. By the theorem of orthogonal complements, it is known that vectors in Co l font size decreased by 3Col A are orthogonal to vectors in Nu l font size decreased by 3Nul ASuperscript Upper TT. Using the definition of orthogonal complements, vectors in Co l font size decreased by 3Col A are orthogonal to vectors in Nu l font size decreased by 3Nul A if and only if the rows and columns of A are the same, which is not necessarily true.
For any scalar c, left norm c Bold v right normcvequals=cleft norm Bold v right normv.
The given statement is false. Since length is always positive, the value of left norm c Bold v right normcv will always be positive. By the same logic, when c is negative, the value of cleft norm Bold v right normv is negative.
For any scalar c, utimes•(cv)equals=c(utimes•v).
The given statement is true because this is a valid property of the inner product.
If left norm Bold u right normusquared2plus+left norm Bold v right normvsquared2equals=left norm Bold u plus Bold v right normu+vsquared2, 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 left norm Bold u plus Bold v right normu+vsquared2equals=left norm Bold u right normusquared2plus+left norm Bold v right normvsquared2.
vtimes•vequals=Bold left norm v right normvsquared2
The given statement is true. By the definition of the length of a vector v, Bold left norm v right normvequals=StartRoot Bold v times Bold v EndRootv•v.
If vectors v11,...,vSubscript pp span a subspace W and if x is orthogonal to each vSubscript jj for jequals=1,...,p, then x is in Upper W Superscript orthogonalW⊥.
The given statement is true. If x is orthogonal to each vSubscript jj, then x is also orthogonal to any linear combination of those vSubscript jj. Since any vector in W can be described as a linear combination of vSubscript jj, x is orthogonal to all vectors in W.
If x is orthogonal to every vector in a subspace W, then x is in Upper W Superscript orthogonalW⊥.
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 is denoted Upper W Superscript orthogonalW⊥.
utimes•vminus−vtimes•uequals=0
The given statement is true. Since the inner product is commutative, utimes•vequals=vtimes•u. Subtracting vtimes•u from each side of this equation gives utimes•vminus−vtimes•uequals=0.
If the columns of an mtimes×n matrix A are orthonormal, then the linear mapping xmaps to↦Ax preserves lengths.
True. left norm Upper A Bold x right norm equals left norm Bold x right normAx=x.
An orthogonal matrix is invertible.
True. An orthogonal matrix is a square invertible matrix U such that Upper U Superscript negative 1 Baseline equals Upper U Superscript Upper TU−1=UT.
Not every orthogonal set in set of real numbers R Superscript nℝn is linearly independent.
True. Every orthogonal set of nonzero vectors is linearly independent, but not every orthogonal set is linearly independent.
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 Bold y equals Bold c 1 Bold u 1 plus times times times plus Bold c Subscript p Baseline Bold u Subscript py=c1u1+•••+cpup can be computed by c Subscript j Baseline equals StartFraction Bold y times Bold u Subscript j Over Bold u Subscript j Baseline times Bold u Subscript j EndFractioncj=y•ujuj•uj, where jequals=1, . . . , p.
Not every linearly independent set in set of real numbers R Superscript nℝn is an orthogonal set.
True. For example, the vectors Start 2 By 1 Table 1st Row 1st Column 0 2nd Row 1st Column 1 EndTable01 and Start 2 By 1 Table 1st Row 1st Column 1 2nd Row 1st Column 1 EndTable11 are linearly independent but not orthogonal.
The orthogonal projection of y onto v is the same as the orthogonal projection of y onto cv whenever c not equals 0c≠0.
True. If c is any nonzero scalar and if v is replaced by cv in the definition of the orthogonal projection of y onto v, then the orthogonal projection of y onto cv is exactly the same as the orthogonal projection of y onto v.
For any two ntimes×n invertible matrices U and V, the inverse of UV is left parenthesis UV right parenthesis Superscript negative 1(UV)−1equals=
V^-1U^-1
To show that left parenthesis UV right parenthesis Superscript negative 1(UV)−1equals=(UV)T, apply the property that states that for matrices U and V with sizes appropriate for multiplication and addition, (UV)Tequals=
V^TU^T
