Determine if U is a subspace of the given vector space V (read describtion)?

V = M(3,3) U= {X E V : X is not invertible}

Thnx for the help in advanced!!

2 Answers

  • 1 decade ago
    Favorite Answer

    It is not a subspace of V. U happens to contain the "zero vector" (the matrix of all 0's), and it is closed under scalar multiplication (if M is not invertible, and t is a scalar, you can prove in any number of ways that tM is not invertible). But U isn't closed under addition!

    Consider the 3x3 matrix with all 0's off the diagonal, and 1, 0, 0 down its main diagonal. Call that matrix A.

    Consider the 3x3 matrix with all 0's off the diagonal, and 0, 1, 1, down its main diagonal. Call that matrix B.

    Note that neither A nor B is invertible. (There are lots of ways to do this: if you like determinants, you can note that since the matrices are diagonal, you can calculate their determinants by taking the product of their diagonal entries, and you get 0 in both cases, so the matrices aren't invertible. You can also show straight from the definitions that neither A nor B is invertible, if that is more to your liking.)

    The sum A+B, on the other hand, is the matrix with all 0s off the diagonal but 1's down the main diagonal. This is the 3x3 identity matrix, often written by I. This _is_ invertible (in fact, I*I = I, so I is its own inverse).

    Since A and B are in U but A+B is not, U is not closed under addition and hence not a vector subspace of V.

    (A and B are not the only examples, of course: you can write down many other noninvertible matrices whose sums are invertible. My A and B are just very simple examples; they come from my first instinct, which is to consider what happens with diagonal matrices before worrying about more general matrices. Often this is very instructive.)

  • Anonymous
    4 years ago

    Checking subspaces is completely approximately constantly rather common. There are 3 tests, and each and each attempt hardly demands extra advantageous proper than 10 seconds concept. The tests are: a million. Is 0 indoors the set S? 2. If x is indoors the S, is a*x indoors the set for any scalar a? 3. If x and y are in S, is x + y in S? case in point, E. enable's do the three tests. a million. Is 0 in S? sure the all 0 matrix is a matrix with det(A) = 0. 2. assume det(A) = 0. Is det(a*A) = 0? sure. when I multiply a matrix via a scalar, I multiply the determinant via a^n. 3. assume det(A) = 0 and det(B) = 0. Does det(A + B) = 0? No. Counter occasion: A = (a million 0) (0 0) B = (0 0) (0 a million) A + B = (a million 0) (0 a million) It passes tests a million and 2 yet fails #3. This set isn't a subspace of Mn(R). See how speedy it extremely is?

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