By Eckhard M.S. Hitzer, University of Fukui, Prof. Ryosuke Nagaoka (University of the Air, Japan)
So please let us have a look at the pattern here. And today I explain to you the sum of subspaces. We have two subspaces U and W of the vector space V. And the sum of the two subspaces U plus W is defined as taking one element from U plus one element y from W and adding them. And so this sum is indeed a new subspace of V. And this can be shown if we look at the criteria of closedness.
Closedness under scalar multiplication is the first one. So if we take an element lambda (times) x plus y, it is lambda x plus lambda y. x is from U, y is from W, and also lambda x is from U and lambda y is from W. So the result is also an element of U plus W. So it is closed under scalar multiplication.
And now we look (under) at the closedness of addition. So we take an element x one plus y one, plus another element x two plus y two. And we can, because it is commutative and associative write it as x one plus x two, plus y one plus y two. And x one plus x two is element of U, y one plus y two is element of W. So the sum of the two again is an element of U plus W. So this new construction, the sum U plus W is now also closed under addition.
So it is closed under scalar multiplication and addition. And this are sufficient criteria for the sum U plus W also to be a real subspace, a vector space which is a subspace of V.
So next I want - please have a gain a look at the pattern - to show to you, that if you have the sum of two subspaces of U plus V, it is not equal the union of the two sets U and V. If we look for example at two straight lines in two dimensions. This be the line U, and this be the line V, then the sum of the two is the subspace, which is the whole space of R 2, here. But the union of them are just all the vectors, here, which are on the two lines. But not any other points. So the union is not equal the sum of the two spaces U and V.
Now I show you an example. We have here three vectors: a one, a two, and a three. And we define the first subspace to be the span of the vectors a one and a two. So it is x a one plus y a two. x and y are elements of (are) real numbers. The second subspace, which we look at is the span of a one and a three. So it is z times a one plus w times a three. z and w are real numbers. And the sum of the two subspaces are all elements from the first subspace, plus the elements from the second subspace. And as you see here you can rewrite them as x plus z times a one, plus y a two, plus w a three. And you see it is not a unique combination. You can take various x and z and it gives you the same, as long as x plus z has the same value, it gives you the same element from R three.
Let us look at another example, where the sum is indeed unique. So here again we have three vectors: a one, a two and a four. And the first subspace shall be the span of a one and a two and the second subspace shall be the span of a four. The sum is taking multiples of a one, plus scalar multiples of a two, so this is the part which comes from U. And the part of the sum which comes from W is z a four. And if we write it in terms of the vectors here explicitly, we see that this combination is indeed unique. It spans the whole space R three and x, y and z are unique for every vector in R three. O.K.
Soli Deo Gloria. Created by
Eckhard Hitzer (Fukui).
Last Modified 16 January 2004
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