Oxford Linear Algebra: Direct Sum of Vector Spaces

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The video begins with the formal mathematical definition of a direct sum via the sum of vector spaces which only have the zero vector in their intersection. This is then shown to be equivalent to an alternative definition which says the representation via the sum of vectors is unique.

Next we look at 3 fully-worked examples where we are asked to check if the given sum of vector spaces is a true direct sum or not. By checking the intersection we see that two of them are not a direct sum as their intersection contains a non-zero vector. For the third example the intersection is shown to be zero, so we then check whether the two subspaces span the larger vector space. By constructing a basis for each subspace, we see that this is indeed the case and thus conclude it is a true direct sum.

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[RCSmiths]

Great information and learning as always Tom. Math rocks!

[samadesh]

Thanks man. You just got another subscriber.

[motherflerkentannhauser]

Here's how I understand this, pls correct me if I'm wrong. There are two types of direct sums (assuming done over finitely many objects):

1. Internal direct sum. This is the idea of, in one single big space V (usually a vector space), splitting V into disjoint subsets s.t. each element from V can be uniquely rewritten as a sum of many components, one from each subset. Like "each sniper combat team consist of a shooter and a spotter". Here x+y=v makes sense since both x, y live in a common space where "+" is defined. In this case you can roughly say "the direct sum of two v-spaces is again a v-space" and the dimensions are added together.

2. External direct sum. This is the same as direct product, cartesian product or tensor product, namely, the ideal of gluing objects that don't have anything to do with each other together and an element in there is an "ordered pair". If each object has its own operation, then the operation of the elements in the sum is done component-wise or independently. Everything stays separate. Take v from V, w from W, you get (v, w) in V x W. You don't do "v+w" since this is undefined and it's not equal to anything. And tensor product is simply adding one more meaning to (v, w), by the universal property: uniquely rewrite it as v \otimes w and define this as a "multilinear map". In this case, "the direct sum of groups/rings/fields/v-spaces" may no longer be another group/ring/field/v-space. You need to carefully test the definitions. And the dimensions are multiplied (because they are independent and it's like counting handshakes or all possible combinations).

[dac]

Could have done with this several years ago in my math undergrad

[ronpearson]

how do you apply linear/abstract algebra regarding direct sums to boolean logic and xnor gates? I know what the tables look like but how would it be worked out mathematically?

[moziburullah]

What is 'direct' about the direct sum? Why cannot we simply call it the sum?

[soutriksaha]

Sir, your looks are absolutely like a are totally opposite of the usual professors of Oxford University in terms of appearance that comes in my totally have proven don't judge a book by it's cover....😊😊😊

[ChiragSonne]

@Tom Rocks Math So, Do you teach GED?

[wolf]

cool ass teacher thx for the knowledge

[mathematicsgurucool]

Hello Sir,

i have learnt about Direct Product also in abstract algebra defined over sets, there they said they need not be abelian groups, but from the book( serge lang undergraduate algebra) i Learnt about this direct sum Tells the two groups must be abelian....why so?please explain

[vansf]

You can have vector sum only when there is a direct physical relationship between 2 different vector. In order to see it more clearly, we can use physics to explain what actually a vector is. It's mathematical representation of a motion, and a motion obviously has a speed, and a direction. Hence a vector = speed + direction . The speed represent the applied force acting on the moving object. When two vectors intersect, the outcome will be the interactions of two different kinetic forces which are acting on the two moving objects, resulting in a 3rd force which deflects the motion of the two objects in two other directions, and the magnitudes of the speeds of the 2 objects and their two other new directions are the very direct sum of space vectors
because each vector alone always is the sum of 3 different vectors pointing toward 3 different directions in the 3-D space

[LucaPasciuta]

That hair is a direct sum of a buzz cut and justin bieber hair

[alicebobson]

youre a term too late!!!! this wouldve helped me so much😭😭😭

[vansf]

if one of the 3 space vectors DNE or is non-existent, then you won't have any 3-D space, instead of that, you will have only a flat surface in human concepts of mathematics, although in objective reality, there is not any actual flat surface, but there are only and always 3-d spaces. Even a single point alone is also a representation of a 3-d space, no matter how extremely small it is because such small points as subatomic particles invisible to human eyes are merely a sort of down-scaled values of visible objects, in terms of the sizes, in the quantum world

[vansf]

The space vector( 1, 1, 1) is called as unit space vector, because other space vectors are merely scalar values of that basic unit space vector, by scaling up to larger or down to smaller than 1. But again, such human invented numbers alone have no actual meaning nor value without any specific and tangible physical world because 1 can be 1 cm or 1 m or 1 km or whatever you want

[vansf]

It is nonsensical or obscure to use mathematical operations to explain why the intersection of two vector must be zero because as mentioned earlier, 0 never exists anywhere in the universe. The only correct way to explain about the intersection part of 2 different vectors is that it is the starting point of the 3rd and resultant vector from the two intersecting vector. It means that the intersection part of the different vector can never be nought or 0, but always have a certain value in a volume space or a 3-D space. The symbolic mathematical value of 0 here can only represents either the equilibrium state of a point in space or a stationary object when all the applied forces acting on it are equal, like in your example where x' = x, y'' = y, and the point or the stationary intersection where the object located can never ever actually be 0 as how your mathematical expressions nonsensically says.
Human defective concepts of mathematics are the very problems causing human fundamental misunderstandings of natural phenomena, and paradoxical theories in all branches of physics and human current development toward self-destruction

[vansf]

Your mathematical language is vague because "space" can be interpreted as a point, a 2 dimensional one, such as plane or flat surface or a 3-D space

[vansf]

Mathematical representations alone without any physical world or any form of matter are both vague and useless, and many times misleading with manipulations of the flaws and defective holes of the concepts of mathematics invented by human limited knowledge. One of the most typical examples is the empty set, which never ever exists anywhere in the universe other than human wild imagination because zero is not any actual value, nor can it represent any form of matter at all, like human -invented concept of infinity

[facts-ecyi]

hey tom, could you try the putnam maths exam? it is one of Americas hardest exams and last i checked only 4 people have ever got a perfect score. It would be really interesting to see how you do on it. Thank you!