# If BC = 0 where B is invertible, show C = 0

• sust0005
In summary, if B is invertible, then B DNE 0, and if B DNE 0 then the only way to have BC = 0 is to have C = 0. However, this is not always true as demonstrated by the example of matrices B and C where BC = 0 but neither B nor C is the zero matrix, showing that the statement "if B DNE 0 then the only way to have BC = 0 is to have C = 0" is untrue. This is because rings of matrices are not always integral domains, meaning that nonzero elements can have a product of zero. This concept can also be seen in other areas of mathematics, such as modular arithmetic.
sust0005
Isn’t it true that if B is invertible then B DNE 0, and if B DNE 0 then the only way to have BC = 0 is to have C = 0?

I'm not seeing (perhaps because I don't want to ;)) how the above is not always true. Can anyone show me an example proving me wrong?

Thanks!

If B is invertible, then B-1 exists.

If you are given that BC = 0, multiply on the left of both sides by B-1. What do you get?

Yes, that I understand (and wish I would've put down on the final!), but what I don't understand is why my initial statement is untrue. That is, why is that not a valid proof that C = 0?

haha, I've seen thru the cloud of trying to convince myself I was correct. My error is now obvious!

Because you didn't use the given information that B is invertible.
For example, consider these matrices B and C.
$$B~=~\left[\begin{array}{c c} 0&1\\ 0&0 \end{array}\right]$$

$$C~=~\left[\begin{array}{c c} 0&2\\ 0&0 \end{array}\right]$$

For these matrices BC = 0, yet neither B nor C is the zero matrix.

Rings of matrices are not in general integral domains (even if the elements are drawn from a field), as Mark44 illustrated with an example. Thus, if you know that BC = 0 and B != 0, you can't conclude that C must be 0 as you would in an integral domain.

This is fairly common, so you should try to keep it in mind. It pops up in a lot of places; matrices are just one example. For instance, the ring of integers modulo 6 is not an integral domain since 2 and 3 are both nonzero but their product is 0 modulo 6. One can also find positive powers of nonzero elements that are zero. Once again using modular arithmetic for an example, the square of two is congruent to 0 modulo 4.

## 1. What is the significance of BC = 0 when B is invertible?

The equation BC = 0 is significant because it suggests that the product of two matrices, B and C, is equal to the zero matrix. This can have implications for solving systems of linear equations and understanding the properties of invertible matrices.

## 2. How does the invertibility of B affect the value of C in the equation BC = 0?

If B is an invertible matrix, then it means that there exists an inverse matrix, B-1, such that B * B-1 = I, where I is the identity matrix. This means that in the equation BC = 0, C must equal the zero matrix in order for the equation to hold.

## 3. Can you provide an example of an invertible matrix B and a corresponding zero matrix C that satisfy the equation BC = 0?

Yes, for example, let B = [1 0; 0 2] and C = [0 0; 0 0]. Both B and C are square matrices of size 2x2, B is invertible, and when multiplied together, the resulting matrix is the zero matrix [0 0; 0 0].

## 4. How can the equation BC = 0 be used to solve systems of linear equations?

If BC = 0, then it can be rewritten as B-1 * BC = B-1 * 0, or C = 0. This can be useful when solving systems of linear equations represented in matrix form, as finding the inverse of B and multiplying it by the resulting zero matrix can help determine the values of the variables in the system.

## 5. Can you prove that C = 0 when BC = 0 and B is invertible?

Yes, since B is an invertible matrix, it has an inverse, B-1, and therefore the equation can be rewritten as B-1 * BC = B-1 * 0. By the properties of matrix multiplication, B-1 * B = I, and I * C = C. Therefore, we have C = B-1 * 0 = 0, proving that C must be equal to the zero matrix.

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